Triangles & Congruence

Study Sheet

Triangles & Congruence

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

Classifying Triangles

Concept
Two ways to sort a triangle
55°

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

By sides:

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

By angles:

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  • 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).

An angle measures a turn.

Example
Example: name a triangle two ways

The triangle below has a square corner mark (a 90∘90^\circ angle) 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.

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.

The Triangle Angle-Sum Theorem

Concept
The angles always add to 180∘180^\circ
55°

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.

An angle measures a turn.

Example
Example: find the missing angle

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}.

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!

The Exterior Angle Theorem

Concept
An outside angle equals two inside angles
55°

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

An angle measures a turn.

Example
Example: use the exterior angle

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}.

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.

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.

Example
Example: read off corresponding parts

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.

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.

Triangle Congruence Shortcuts

Concept
Five ways to prove congruence
55°

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

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  • 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.

An angle measures a turn.

Example
Example: a congruent pair marked with SAS

Matching tick marks show equal sides; the arc shows the equal included angle. Two sides and the angle between them match, so this is SAS.

Therefore △ABC≅△DEF\triangle ABC \cong \triangle DEF by SAS.

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.

Isosceles and Equilateral Triangles

Concept
Base Angles Theorem
55°

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.

An angle measures a turn.

Example
Example: find the vertex angle

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}.

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).

The Triangle Inequality

Concept
Can these lengths make a triangle?
55°

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.

An angle measures a turn.

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.
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.

A First Two-Column Proof

Concept
What a proof looks like
55°

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.

An angle measures a turn.

Example
Example: prove two triangles congruent

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

Notice the pattern: Side (ABAB), Angle (at BB), Side (BMBM) --- that is SAS!

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.

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.

An angle measures a turn.

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).

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  • 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.
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

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.

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:

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  • 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

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

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 whole base is BC=14BC = 14. 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 1414 splits into 4+3=74+3 = 7 equal shares, each of length 14/7=214/7 = 2:

BD=4×2=8,DC=3×2=6.BD = 4\times 2 = \boxed{8}, \qquad DC = 3\times 2 = \boxed{6}.

Check: 8+6=148 + 6 = 14, and 8:6=4:38:6 = 4:3 as required.

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

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.

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

Concept
The full triangle-inequality range

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.

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.