Triangles & Congruence

Study Sheet

Triangles & Congruence

Classifying, angle rules, congruence shortcuts, and a first proof

Classifying Triangles

Concept
Two ways to sort a triangle

Every triangle can be described in two ways at once: by its sides and by its angles.

By sides:

  • Scalene — all three sides different lengths.
  • Isosceles — at least two sides equal.
  • Equilateral — all three sides equal.

By angles:

  • Acute — all three angles less than 90∘90^\circ.
  • Right — exactly one 90∘90^\circ angle.
  • Obtuse — one angle greater than 90∘90^\circ.
  • Equiangular — all three angles equal (60∘60^\circ each).

Two triangles are congruent when they match exactly, and you never need all six parts to know it: SSS, SAS, ASA, AAS (and HL for right triangles) each suffice. Once triangles are proved congruent, every remaining pair of corresponding parts is equal — CPCTC — which is how most geometry proofs finish.

Example
Example: name a triangle two ways
ABC
One tick on each short side: the legs are equal.

The triangle in the figure has a square corner mark (a 90∘90^\circ angle) at AA, and its two short sides carry matching tick marks, meaning they are equal.

By sides: two equal sides ⇒\Rightarrow isosceles.   By angles: one 90∘90^\circ angle ⇒\Rightarrow right.

So this is an isosceles right triangle; its other two angles are equal and share the remaining 90∘90^\circ, so each is 45∘45^\circ.

Side note
Where it goes wrong

Using SSA (two sides and a non-included angle) as a congruence test. It fails: the same data can build two different triangles. Only HL, the right-angle case of SSA, works.

Tip

Tip: “Equilateral” and “equiangular” are partners — if all three sides are equal, then all three angles are equal (60∘60^\circ), and the other way around too.

Try it
Quick check

A triangle has sides 77, 77 and 727\sqrt2. Classify it by its sides and by its angles.

Show the answer

Two equal sides: isosceles. For the angles compare squares: 72+72=987^2 + 7^2 = 98 and (72)2=49⋅2=98(7\sqrt2)^2 = 49 \cdot 2 = 98. They are equal, so by the converse of the Pythagorean Theorem the triangle is right — an isosceles right triangle.

The Triangle Angle-Sum Theorem

Concept
The angles always add to 180∘180^\circ

In any triangle, the three interior angles add up to exactly 180∘180^\circ.

∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^\circ

If you know two angles, subtract their sum from 180∘180^\circ to find the third.

This is a theorem, not an observation — measuring a thousand triangles would never prove it. The proof, in the box below, uses one parallel line and the alternate interior angles it creates. It also explains the corollaries at a glance: a triangle has at most one right or obtuse angle, and the two acute angles of a right triangle are complementary.

Theorem
Why the sum is always 180∘180^\circ
ababcABC
The parallel through C carries angles A and B up to C.

Draw the line through CC parallel to AB‾\overline{AB}. With AC‾\overline{AC} as a transversal, the angle at CC on the left is an alternate interior angle to ∠A\angle A, so it equals aa. With BC‾\overline{BC} as a transversal, the angle on the right equals ∠B=b\angle B = b. Those two angles and ∠C\angle C sit side by side along a straight line, so

a+c+b=180∘.a + c + b = 180^\circ .
Example
Example: find the missing angle
65°40°xABC
Drawn to scale: the missing angle at C is 75∘^\circ.

Two angles of a triangle are 65∘65^\circ and 40∘40^\circ. Find the third angle xx.

Add the two known angles: 65∘+40∘=105∘65^\circ + 40^\circ = 105^\circ. Then x=180∘−105∘=75∘x = 180^\circ - 105^\circ = \boxed{75^\circ}.

Side note
Where it goes wrong

Measuring instead of computing. A drawing that is not marked "to scale" may show a 75∘75^\circ angle as 60∘60^\circ; the theorem, not the ruler, gives the answer.

Tip

Tip: A triangle can have at most one right or obtuse angle. Two right angles alone would already use up 180∘180^\circ, leaving nothing for the third!

Try it
Quick check

The angles of a triangle are xx, 2x2x and 3x3x. Find them and classify the triangle by its angles.

Show the answer

x+2x+3x=180x + 2x + 3x = 180, so 6x=1806x = 180 and x=30x = 30. The angles are 30∘30^\circ, 60∘60^\circ, 90∘90^\circ: a right triangle.

The Exterior Angle Theorem

Concept
An outside angle equals two inside angles

If you extend one side of a triangle, you create an exterior angle. That exterior angle equals the sum of the two remote (far-away) interior angles.

exterior angle=remote angle1+remote angle2\text{exterior angle} = \text{remote angle}_1 + \text{remote angle}_2

Why: the exterior angle yy and the interior angle beside it form a linear pair, so y=180∘−∠By = 180^\circ - \angle B. The angle-sum theorem says ∠A+∠C=180∘−∠B\angle A + \angle C = 180^\circ - \angle B too. Both equal the same thing, so y=∠A+∠Cy = \angle A + \angle C. A useful corollary: an exterior angle is larger than either remote interior angle — the fact behind the proof that equal alternate interior angles force parallel lines.

Example
Example: use the exterior angle
70°50°yABCD
Side AB is extended to D; y is outside the triangle at B.

Side ABAB is extended to point DD. The two remote interior angles are A=70∘A=70^\circ and C=50∘C=50^\circ.

The exterior angle y=70∘+50∘=120∘y = 70^\circ + 50^\circ = \boxed{120^\circ}.

Side note
Where it goes wrong

Adding the wrong pair. The exterior angle at BB equals the two angles not at BB. Adding the interior angle at BB (60∘60^\circ) to one remote angle gives nonsense; the interior angle at BB is the exterior angle's supplement.

Tip

Tip: The exterior angle and its neighbor interior angle always add to 180∘180^\circ (a straight line). That is a handy way to check your work.

Try it
Quick check

An exterior angle of a triangle measures 130∘130^\circ. The two remote interior angles are (2x+10)∘(2x + 10)^\circ and 3x∘3x^\circ. Find them.

Show the answer

Exterior angle == sum of the remote interior angles: (2x+10)+3x=130(2x + 10) + 3x = 130, so 5x=1205x = 120 and x=24x = 24. The angles are 2(24)+10=58∘2(24) + 10 = \mathbf{58^\circ} and 3(24)=72∘3(24) = \mathbf{72^\circ} (and the third angle is 180−130=50∘180 - 130 = 50^\circ).

Congruent Figures and Corresponding Parts

Concept
Same shape, same size

Two figures are congruent (≅\cong) if they have exactly the same shape and size — one could be slid, turned, or flipped onto the other perfectly. When we write △ABC≅△DEF\triangle ABC \cong \triangle DEF, the order of the letters tells us which parts match:

A↔D,B↔E,C↔F.A\leftrightarrow D,\quad B\leftrightarrow E,\quad C\leftrightarrow F.

These matching pieces are called corresponding parts.

"Congruent" is defined by motion: one figure can be carried onto the other by slides, turns and flips, which never change a length or an angle. The motion pairs up the vertices, and the order of letters in △ABC≅△DEF\triangle ABC \cong \triangle DEF records that pairing. That is why the order is not decoration — a different order is a different (and usually false) claim.

Example
Example: read off corresponding parts
99ABCDEF
DEF is a mirror image of ABC; matching ticks mark corresponding sides.

Given △ABC≅△DEF\triangle ABC \cong \triangle DEF, the matching sides and angles are:

AB‾≅DE‾,BC‾≅EF‾,AC‾≅DF‾,\overline{AB}\cong\overline{DE},\quad \overline{BC}\cong\overline{EF},\quad \overline{AC}\cong\overline{DF},
∠A≅∠D,∠B≅∠E,∠C≅∠F.\angle A\cong\angle D,\quad \angle B\cong\angle E,\quad \angle C\cong\angle F.

So if BC=9BC = 9, then EF=9EF = 9 too, because they correspond.

Side note
Where it goes wrong

Matching parts by position on the page instead of by the letters. In the figure, △DEF\triangle DEF is flipped, so EE is on the left; pairing "bottom-left with bottom-left" would match AA with EE, which the statement △ABC≅△DEF\triangle ABC \cong \triangle DEF does not say.

Tip

CPCTC stands for “Corresponding Parts of Congruent Triangles are Congruent.” Once you prove two triangles congruent, every pair of matching parts is automatically congruent.

Try it
Quick check

△PQR≅△XYZ\triangle PQR \cong \triangle XYZ, with PQ=8PQ = 8, QR=11QR = 11 and m∠R=40∘m\angle R = 40^\circ. Which side of △XYZ\triangle XYZ has length 1111, and which of its angles measures 40∘40^\circ?

Show the answer

Read the pairing from the letter order: P↔XP \leftrightarrow X, Q↔YQ \leftrightarrow Y, R↔ZR \leftrightarrow Z. So QR‾\overline{QR} corresponds to YZ‾\overline{YZ}, giving YZ=11\mathbf{YZ = 11}, and ∠R\angle R corresponds to ∠Z=40∘\mathbf{\angle Z = 40^\circ}.

Triangle Congruence Shortcuts

Concept
Five ways to prove congruence

You do not need to check all six parts (3 sides + 3 angles). Any one of these shortcuts is enough:

  • SSS — three pairs of sides.
  • SAS — two sides and the included angle (the angle between them).
  • ASA — two angles and the included side.
  • AAS — two angles and a non-included side.
  • HL — (right triangles only) the hypotenuse and one leg.

Each shortcut works because its three parts already pin the triangle down: once you know them you could build only one triangle. With SAS, the two sides and the angle between them fix all three vertices; with ASA, the two angles fix the directions of the other two sides from the ends of the known side, and two directions from two points meet in exactly one place. AAS reduces to ASA, since two angles give the third. HL is the right-angle case where the Pythagorean Theorem supplies the third side.

Example
Example: a congruent pair marked with SAS
ABCDEF
Sides 5 and 7 with a 50∘^\circ angle between them, in both triangles.

Matching tick marks show equal sides: AB=DEAB = DE (one tick) and BC=EFBC = EF (two ticks). The arcs show ∠B=∠E\angle B = \angle E, and each of those angles sits between the two marked sides.

Two sides and the included angle match, so △ABC≅△DEF\triangle ABC \cong \triangle DEF by SAS — even though △DEF\triangle DEF is turned on the page.

Side note
Where it goes wrong

Using SSA. Two sides and an angle that is not between them can fit two different triangles: with ∠A=30∘\angle A = 30^\circ, AB=10AB = 10 and BC=6BC = 6, the side BCBC can swing to meet the base at two different points (see "Why It Works" below). Only when the angle is right (HL) is SSA safe.

Tip

The two that do NOT work: SSA (two sides and a non-included angle) is unreliable — it can produce two different triangles (the “ambiguous case”). AAA (three angles) only guarantees the same shape, not the same size — think of a small triangle and a giant one with equal angles.

Try it
Quick check

Which shortcut, if any, proves △ABC≅△DEF\triangle ABC \cong \triangle DEF?

  • AB=DEAB = DE, BC=EFBC = EF, ∠B=∠E\angle B = \angle E
  • ∠A=∠D\angle A = \angle D, ∠B=∠E\angle B = \angle E, BC=EFBC = EF
  • AB=DEAB = DE, BC=EFBC = EF, ∠A=∠D\angle A = \angle D
Show the answer

(1) ∠B\angle B is between AB‾\overline{AB} and BC‾\overline{BC}: SAS. (2) Two angles and a side that is not between them (BC‾\overline{BC} is opposite ∠A\angle A): AAS. (3) ∠A\angle A is not between AB‾\overline{AB} and BC‾\overline{BC}: this is SSA, which is not enough.

Isosceles and Equilateral Triangles

Concept
Base Angles Theorem

In an isosceles triangle, the two sides that are equal are the legs, and the third side is the base. The theorem says:

The angles opposite the equal sides (the base angles) are congruent.

And the reverse is true too: if two angles are equal, the sides opposite them are equal.

Why: draw the bisector of the vertex angle down to the base. It splits the triangle into two triangles with a pair of equal legs, equal halves of the vertex angle, and a shared side — congruent by SAS — so the base angles, which correspond, are equal. (The full two-column version is in Going Deeper.) The same picture shows the bisector is also the median and the altitude: it hits the base at its midpoint, at a right angle.

Example
Example: find the vertex angle
50°50°80°
Equal legs (ticks) face equal base angles.

The two legs are marked equal, so the two base angles are equal (50∘50^\circ each).

Vertex angle =180∘−(50∘+50∘)=80∘= 180^\circ - (50^\circ + 50^\circ) = \boxed{80^\circ}.

Side note
Where it goes wrong

Making the wrong angles equal. The equal angles are the ones opposite the equal sides — the base angles — not the angle between the equal sides. If a problem says a vertex angle is 50∘50^\circ, the base angles are 180−502=65∘\tfrac{180 - 50}{2} = 65^\circ each, not 50∘50^\circ.

Tip

Equilateral facts: every equilateral triangle is also isosceles, is equiangular, and has three 60∘60^\circ angles. Its perimeter is simply 3×3 \times (one side).

Try it
Quick check

The vertex angle of an isosceles triangle measures 36∘36^\circ. Find each base angle.

Show the answer

The base angles are equal and share 180∘−36∘=144∘180^\circ - 36^\circ = 144^\circ, so each is 72∘\mathbf{72^\circ}.

The Triangle Inequality

Concept
Can these lengths make a triangle?
AB236
Sides 2 and 3 swing on arcs from the ends of 6, and the arcs never meet.

Three lengths form a triangle only if the two shorter ones can “reach across.” The rule:

The sum of any two sides must be greater than the third side.

A quick check: add the two smallest sides. If that sum beats the largest side, you have a triangle.

The picture is the proof. The third vertex must be 22 from AA and 33 from BB, so it lies on both dashed arcs. Those arcs reach only 22 and 33 along the base, leaving a gap of 11 in the middle: they never meet, and no triangle exists. If the two short sides added to exactly 66 the arcs would touch on the base, giving a flat "triangle". The rule is the geometric form of "the straight path is the shortest": going from AA to BB through a third point is always longer than going straight.

Example
Example: test some lengths

Lengths 5,7,95, 7, 9: smallest two are 5+7=125+7=12, and 12>912 > 9. Yes, a triangle.

Lengths 2,3,62, 3, 6: smallest two are 2+3=52+3=5, but 5<65 < 6. No triangle — the short sides cannot reach.

Range for a third side: if two sides are 88 and 55, the third side ss must satisfy

8−5<s<8+5,that is3<s<13.8-5 < s < 8+5, \quad\text{that is}\quad 3 < s < 13.
Side note
Where it goes wrong

Checking only one inequality with the wrong sides. For 3,4,103, 4, 10: 3+10>43 + 10 > 4 and 4+10>34 + 10 > 3 are true, but 3+4>103 + 4 > 10 fails. The test that can fail is always "two shortest versus longest".

Tip

Tip: The third side is always between the difference and the sum of the other two:   ∣a−b∣<s<a+b\;|a-b| < s < a+b.

Try it
Quick check

Two sides of a triangle are 44 and 1111. What values can the third side take? Is 77 possible?

Show the answer

11−4<s<11+411 - 4 < s < 11 + 4, so 7<s<15\mathbf{7 < s < 15}. s=7s = 7 is not possible: 4+7=114 + 7 = 11 exactly, so the "triangle" would lie flat.

A First Two-Column Proof

Concept
What a proof looks like

A two-column proof lists Statements on the left and the Reason for each on the right. You start from what is Given and end at what you want to Prove, one logical step at a time.

A proof is a chain in which every link is either given or follows from earlier links by a definition, postulate or theorem you can name. The two columns keep that honest: a statement with no reason, or a reason that does not apply, is visible at a glance. The last statement is the one you were asked to prove.

Example
Example: prove two triangles congruent
ACMB
Given: AB ≅\cong CB and BM bisects angle ABC.

Given: AB‾≅CB‾\overline{AB}\cong\overline{CB}, and BM‾\overline{BM} bisects ∠ABC\angle ABC.   Prove: △ABM≅△CBM\triangle ABM \cong \triangle CBM.

StatementReason1. AB‾≅CB‾Given2. ∠ABM≅∠CBMDefinition of angle bisector3. BM‾≅BM‾Reflexive Property4. △ABM≅△CBMSAS (steps 1, 2, 3)\begin{array}{l|l} \textbf{Statement} & \textbf{Reason} \\ \hline 1.\ \overline{AB}\cong\overline{CB} & \text{Given} \\ 2.\ \angle ABM \cong \angle CBM & \text{Definition of angle bisector} \\ 3.\ \overline{BM}\cong\overline{BM} & \text{Reflexive Property} \\ 4.\ \triangle ABM \cong \triangle CBM & \text{SAS (steps 1, 2, 3)} \end{array}

Notice the pattern: Side (ABAB), Angle (at BB), Side (BMBM) — that is SAS, and the angle really is between the two sides.

Side note
Where it goes wrong

Writing "AM=CMAM = CM" as a step. It is true, but it is what congruence gives (by CPCTC after step 4), not something you may use before. Using a consequence of the conclusion as a reason is circular.

Tip

Tip: The Reflexive Property (BM‾≅BM‾\overline{BM}\cong\overline{BM}) is your best friend in proofs. Whenever two triangles share a side or an angle, that shared part is congruent to itself.

Try it
Quick check

Given AB‾≅CB‾\overline{AB} \cong \overline{CB} and AM‾≅CM‾\overline{AM} \cong \overline{CM} in the same figure. Which postulate proves △ABM≅△CBM\triangle ABM \cong \triangle CBM, and what is the third pair of parts?

Show the answer

The third pair is the shared side BM‾≅BM‾\overline{BM} \cong \overline{BM} (Reflexive Property). Three pairs of sides: SSS. Then CPCTC gives ∠ABM≅∠CBM\angle ABM \cong \angle CBM — so BM‾\overline{BM} bisects ∠ABC\angle ABC.

Going Deeper: Advanced Triangle Ideas

You have met the everyday rules of triangles. Now we push further into the ideas that professional geometers actually use — special points inside a triangle, a clever “inner” triangle, a theorem that slices sides in a neat ratio, proofs that add a line of their own, and the full story of which lengths can form a triangle. Sketch along and enjoy the extra depth.

Concept
The four centers of a triangle

Every triangle has four famous “centers,” each built from a different set of special lines. All three lines of a kind always meet at a single point (they are concurrent).

  • Centroid — where the three medians meet (a median joins a vertex to the midpoint of the opposite side). It is the triangle's balance point, and it always sits inside.
  • Incenter — where the three angle bisectors meet. It is the center of the inscribed circle that just touches all three sides, and it is always inside.
  • Circumcenter — where the three perpendicular bisectors of the sides meet. It is the center of the circle through all three vertices; it may fall outside for an obtuse triangle.
  • Orthocenter — where the three altitudes (perpendicular heights from each vertex) meet. It too can land outside an obtuse triangle.

Why the lines of each kind meet at one point: for the perpendicular bisectors, the bisector of AB‾\overline{AB} is every point equidistant from AA and BB, and the bisector of BC‾\overline{BC} is every point equidistant from BB and CC; where they cross, the point is equidistant from all three vertices, so it is on the third bisector too. The angle bisectors work the same way with distances to the sides. The medians meet because each median divides the triangle's "balance" evenly, and the centroid's 2:12:1 split comes from the midsegment joining the feet of two medians being parallel to, and half of, the third side.

Tip

The centroid's 2:1 rule: the centroid divides each median so that the piece from the vertex is twice as long as the piece to the midpoint. So the vertex-to-centroid part is 23\tfrac{2}{3} of the whole median, and the centroid-to-midpoint part is 13\tfrac{1}{3}.

Example
Example: use the centroid's 2:1 ratio
ABCMG126
All three medians pass through G, which sits two-thirds of the way down each one.

In △ABC\triangle ABC below, MM is the midpoint of BC‾\overline{BC}, so AM‾\overline{AM} is a median. The three medians meet at the centroid GG. Suppose the whole median AM=18AM = 18. Find AGAG and GMGM.

The centroid splits the median in the ratio 2:12:1 from the vertex. So the vertex part is 23\tfrac{2}{3} of 1818 and the midpoint part is 13\tfrac{1}{3} of 1818:

AG=23×18=12,GM=13×18=6.AG = \tfrac{2}{3}\times 18 = \boxed{12}, \qquad GM = \tfrac{1}{3}\times 18 = \boxed{6}.

Check: 12+6=1812 + 6 = 18, and 1212 is indeed twice 66.

Side note
Where it goes wrong

Reversing the 2:12 : 1 split. The longer piece is next to the vertex: with AM=18AM = 18, AG=12AG = 12 and GM=6GM = 6, not AG=6AG = 6. The centroid sits closer to the side than to the vertex.

Concept
The medial triangle and the Midsegment Theorem

Join the midpoints of the three sides of a triangle and you get a smaller triangle inside, called the medial triangle. Each of its sides is a midsegment of the original. The Midsegment Theorem says a midsegment is:

  • parallel to the third side of the triangle, and
  • exactly half as long as that side.

Because all three sides are halved, the medial triangle has the same shape as the original, half the perimeter, and one quarter of the area.

Example
Example: a midsegment length
105ABCPQ
PQ joins two midpoints: parallel to BC and half as long.

PP, QQ, RR are the midpoints of the sides of △ABC\triangle ABC. Side BC‾\overline{BC} has length 1010. Find the midsegment PQ‾\overline{PQ} that joins the midpoints of AB‾\overline{AB} and AC‾\overline{AC}.

PQ‾\overline{PQ} joins the midpoints of the two sides meeting at AA, so it is the midsegment parallel to BC‾\overline{BC}. By the Midsegment Theorem it is half of BCBC:

PQ=12×10=5,andPQ‾∥BC‾.PQ = \tfrac{1}{2}\times 10 = \boxed{5}, \quad\text{and}\quad \overline{PQ}\parallel\overline{BC}.
Concept
The Angle Bisector Theorem

Draw the bisector of one angle of a triangle down to the opposite side. It cuts that side into two pieces whose lengths are in the same ratio as the two sides of the angle. In △ABC\triangle ABC, if the bisector from AA meets BC‾\overline{BC} at DD, then

BDDC=ABAC.\frac{BD}{DC} = \frac{AB}{AC}.

The bisector divides the far side in proportion to the two nearby sides.

Example
Example: split a side with the Angle Bisector Theorem
8643ABCD
The bisector from A cuts BC in the ratio 8 : 6.

In △ABC\triangle ABC, the bisector from AA meets BC‾\overline{BC} at DD. The two sides are AB=8AB = 8 and AC=6AC = 6, and the base is BC=7BC = 7. Find BDBD and DCDC.

By the theorem, BDDC=ABAC=86=43\dfrac{BD}{DC} = \dfrac{AB}{AC} = \dfrac{8}{6} = \dfrac{4}{3}. So the base of 77 splits into 4+3=74+3 = 7 equal shares, each of length 7/7=17/7 = 1:

BD=4×1=4,DC=3×1=3.BD = 4\times 1 = \boxed{4}, \qquad DC = 3\times 1 = \boxed{3}.

Check: 4+3=74 + 3 = 7, and 4:3=8:64:3 = 8:6 as required. (Sides 88, 66, 1414 would not do: 8+6=148 + 6 = 14, so that "triangle" would be flat.)

Concept
Congruence proofs with an auxiliary line

Sometimes a proof seems stuck because there are not yet two triangles to compare. The trick is to add your own line — called an auxiliary line — that creates the triangles you need. Popular choices are an angle bisector, a median, an altitude, or a segment joining two existing points. Draw it, justify why it exists, and the congruence shortcuts (SSS, SAS, ASA, AAS, HL) do the rest.

Example
Example: prove the base angles of an isosceles triangle are equal
ABCD
The dashed bisector AD is the auxiliary line.

Given: △ABC\triangle ABC with AB‾≅AC‾\overline{AB}\cong\overline{AC}.   Prove: ∠B≅∠C\angle B \cong \angle C.

Auxiliary line: draw the bisector of ∠A\angle A, meeting BC‾\overline{BC} at DD. This single extra segment splits the triangle into two triangles we can compare.

StatementReason1. AB‾≅AC‾Given2. Draw bisector AD‾ of ∠AEvery angle has a bisector3. ∠BAD≅∠CADDefinition of angle bisector4. AD‾≅AD‾Reflexive Property5. △ABD≅△ACDSAS (steps 1, 3, 4)6. ∠B≅∠CCPCTC\begin{array}{l|l} \textbf{Statement} & \textbf{Reason} \\ \hline 1.\ \overline{AB}\cong\overline{AC} & \text{Given} \\ 2.\ \text{Draw bisector } \overline{AD} \text{ of } \angle A & \text{Every angle has a bisector} \\ 3.\ \angle BAD \cong \angle CAD & \text{Definition of angle bisector} \\ 4.\ \overline{AD}\cong\overline{AD} & \text{Reflexive Property} \\ 5.\ \triangle ABD \cong \triangle ACD & \text{SAS (steps 1, 3, 4)} \\ 6.\ \angle B \cong \angle C & \text{CPCTC} \end{array}

The auxiliary bisector turned one triangle into two matching ones, and CPCTC finished the job.

Concept
The full triangle-inequality range
85
Hinge the 5 at one end of the 8: the dashed third side grows from near 3 to near 13 as the hinge opens.

For three sides aa, bb, cc all three sums must beat the remaining side:

a+b>c,b+c>a,a+c>b.a+b>c,\qquad b+c>a,\qquad a+c>b.

Turned around, if two sides are aa and bb, the third side ss is squeezed between their difference and their sum:

∣a−b∣  <  s  <  a+b.|a-b| \;<\; s \;<\; a+b.

Both ends are strict — landing exactly on ∣a−b∣|a-b| or a+ba+b would flatten the triangle into a straight line.

Example
Example: count the integer third sides

Two sides of a triangle are 88 and 55. How many whole-number lengths are possible for the third side ss?

Apply the range with a=8a=8, b=5b=5:

∣8−5∣<s<8+5⟹3<s<13.|8-5| < s < 8+5 \quad\Longrightarrow\quad 3 < s < 13.

Since the ends are strict, ss cannot equal 33 or 1313. The allowed integers are

s∈{4,5,6,7,8,9,10,11,12}.s \in \{4,5,6,7,8,9,10,11,12\}.

Counting them (or using 12−4+112 - 4 + 1) gives 9\boxed{9} possible integer lengths.

Tip

Exterior angles in an angle chase: in a longer figure, keep reusing “exterior angle == sum of the two remote interior angles.” Each time you meet an extended side, you can jump straight to a new angle without first finding the interior one — a fast shortcut when hunting an unknown angle several steps away.

Try it
Quick check

GG is the centroid of △ABC\triangle ABC and MM is the midpoint of BC‾\overline{BC}. If GM=7GM = 7, find AGAG and the median AMAM.

Show the answer

GMGM is the short third of the median, so AM=3⋅7=21AM = 3 \cdot 7 = \mathbf{21} and AG=2⋅7=14AG = 2 \cdot 7 = \mathbf{14}.

Why It Works

Theorem
The reason behind the rule
30°1066ABC₁C₂
SSA: angle 30∘^\circ, sides 10 and 6 — and two different triangles ABC1_1, ABC2_2.

Why SAS is enough: place the two triangles so the equal angles coincide and one pair of equal sides lies along the same ray. The second pair of equal sides must then lie along the other ray of the angle, ending at the same point. Three vertices now coincide, so the triangles coincide. SSA fails because the second side's length pins its endpoint to a circle, and a circle can meet a ray twice — two triangles, one set of data.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
The Pythagorean theorem
a2+b2=c2a^2+b^2=c^2

What it means. In a right triangle the squares on the legs add to the square on the hypotenuse.

Example. Legs 33 and 44: c=32+42=5c=\sqrt{3^2+4^2}=5.

Why it works. Take four copies of the triangle and place them inside a square of side a+ba+b, leaving a tilted square of side cc in the middle. The big square's area is (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2; it is also the four triangles 4⋅12ab=2ab4\cdot\tfrac12 ab=2ab plus c2c^2. Cancelling 2ab2ab leaves a2+b2=c2a^2+b^2=c^2.

Tip. cc must be the hypotenuse — the side opposite the right angle, always the longest. The converse also holds: if a2+b2=c2a^2+b^2=c^2 the triangle is right-angled.

Tip
Triangle angle sum and the exterior angle
A+B+C=180∘,exterior=sum of the two remote interior anglesA+B+C=180^\circ, \qquad \text{exterior} = \text{sum of the two remote interior angles}

What it means. Every triangle's angles total a straight angle; an exterior angle equals the two angles it is not next to.

Example. Angles 50∘50^\circ and 60∘60^\circ give a third of 70∘70^\circ; the exterior angle there is 110∘110^\circ.

Why it works. Draw a line through one vertex parallel to the opposite side. The two alternate interior angles equal the other two triangle angles, and together with the third they form a straight line — 180∘180^\circ. The exterior result follows since exterior =180∘−adjacent=180^\circ-\text{adjacent}, and the other two also total 180∘−adjacent180^\circ-\text{adjacent}.

Tip. The exterior angle shortcut saves a step: no need to find the third angle first.

Tip
Heron's formula
A=s(s−a)(s−b)(s−c),s=a+b+c2A=\sqrt{s(s-a)(s-b)(s-c)},\qquad s=\frac{a+b+c}{2}

What it means. The area of a triangle from its three sides alone — no angle or height needed.

Example. Sides 3,4,53,4,5: s=6s=6, so A=6⋅3⋅2⋅1=6A=\sqrt{6\cdot 3\cdot 2\cdot 1}=6.

Why it works. Start from A=12absin⁡CA=\tfrac12 ab\sin C, replace sin⁡C\sin C with 1−cos⁡2C\sqrt{1-\cos^2C}, and substitute cos⁡C\cos C from the Law of Cosines. The algebra factors into the four bracketed terms.

Tip. ss is the SEMI-perimeter — half the perimeter. Forgetting the halving is the usual slip.

Tip
The Triangle Inequality
a+b>cfor every pair of sidesa+b>c \quad\text{for every pair of sides}

What it means. Any two sides of a triangle must together exceed the third.

Example. Sides 2,3,62,3,6 cannot form a triangle since 2+3<62+3<6.

Why it works. The straight path between two vertices is the shortest one, so going via the third vertex can only be longer.

Tip. To test whether three lengths form a triangle, just check the two shortest against the longest.