Triangles, congruence, and similarity

Learning objectives

Use triangle angle and side relationships; identify which information guarantees congruence or similarity; preserve vertex correspondence; set up consistent proportions; distinguish length scaling from area scaling.

Triangle structure: what must be true

Every nondegenerate triangle has three positive interior angles totaling 180°. An isosceles triangle has at least two equal sides; the angles opposite those sides are equal. Conversely, equal angles have equal opposite sides. An equilateral triangle has three equal sides and three 60° angles. Match a side to the angle opposite it. In △ABC, side AB is opposite ∠C, side BC is opposite ∠A, and side AC is opposite ∠B. The largest angle is opposite the longest side. The word “opposite” means the side that does not touch the angle’s vertex.

For positive side lengths a, b, c, the triangle inequality requires each side to be less than the sum of the other two. If a and b are fixed, the possible third side satisfies

$|a - b| < c < a + b.$

The inequalities are strict: equality would collapse the triangle onto a line. When a problem asks for integer possibilities, find the open interval first, then count the integers inside it.

The equal angles are opposite the marked equal sides. The vertex angle at A need not equal either base angle.

Congruence and similarity are different claims

Congruent triangles have the same shape and size. Their corresponding sides and angles are equal, even if one triangle has been rotated, translated, or reflected.

Similar triangles have the same shape, but not necessarily the same size. Corresponding angles are equal; corresponding side lengths have one common positive ratio.

Claim	Sufficient information	Important qualification
Congruence	SSS; SAS; ASA; AAS	SAS uses the angle between the two given sides. ASA/AAS use two angles and a corresponding side.
Congruence of right triangles	Hypotenuse–leg (HL)	Both triangles must be right triangles; the equal hypotenuses and one pair of equal legs suffice.
Similarity	AA; proportional SSS; proportional SAS	In SAS similarity, the included angles must be equal and the two pairs of sides must share a ratio.

Here S means side and A means angle. AA is enough for similarity because the third angles automatically match. AAA alone is not enough for congruence: all equilateral triangles have the same angles, but their sizes can differ. SSA, two sides and a nonincluded angle, is not a general congruence criterion; in some arrangements it allows two different triangles.

Correspondence comes before calculation

The statement △ABC ∼ △DEF specifies

A ↔ D, B ↔ E, C ↔ F,

so AB ↔ DE, BC ↔ EF, and AC ↔ DF. This order remains true even when the figures are drawn with different orientations.

If the scale factor from the first triangle to the second is k, then

$\frac{DE}{AB}$

Write all ratios in the same direction. For example, “large over small” must remain large over small in every fraction. Alternatively, equate within-triangle ratios such as $AB/BC = DE/EF.$ Both approaches work; mixing them does not.

Parallel segments and scaling

If D is on AB, E is on AC, and $DE ∥ BC,$ then △ADE ∼ △ABC by angle-angle similarity. Thus

$\frac{AD}{AB}$

The denominator uses the whole side, not automatically the remaining piece. Since $AB = AD + DB$ and $AC = AE + EC,$ a part-to-whole setup is often the safest starting point. The derived part-to-part equality $AD/DB = AE/EC$ is also valid in this parallel-segment configuration.

When similar figures scale by k, all corresponding lengths scale by k: sides, perimeters, altitudes, medians, and radii. Areas scale by k² because both base and height scale:

$\frac{1}{2}$

Angles do not change. Conversely, an area ratio of q produces a length ratio of $\sqrt{q}$ not q.

Same height is useful even without similarity

Triangles sharing the same altitude have areas in the ratio of their bases, because the common factor $h/2$ cancels. This is a different argument from k² scaling: do not square a base ratio unless the whole triangles are similar.

15 worked examples

2.01. Angles given as a ratio

A triangle’s interior angles are in the ratio 2 : 3 : 4. What is the measure of its largest angle?

Show worked solutionHide worked solution

Recognize the structure. A ratio gives relative parts, so introduce one common multiplier.

Work it through. Write the angles as 2t, 3t, and 4t degrees. The triangle sum gives

$2t + 3t + 4t = 180 9t = 180 t = 20.$

The angles are 40°, 60°, and 80°.

Answer: 80°.

Key takeaway. The ratio entries are not angle measures. They tell you how the fixed total is partitioned. This “common multiplier” method also works for side lengths and perimeters.

2.02. An isosceles triangle’s base angles

In △ABC, AB = AC and ∠A = 44°. What is the measure of ∠B?

Show worked solutionHide worked solution

Recognize the structure. The equal sides meet at A, so B and C are the equal base angles.

Work it through. Let each base angle be b degrees. Then $44 + 2b = 180,$ so $2b = 136$ and $b = 68.$

Answer: 68°.

Avoid the trap. The 44° angle is the vertex angle between the equal sides, not one of the equal base angles. Identify the opposite sides and angles explicitly.

2.03. Equal angles determine equal sides

In △ABC, ∠B = ∠C. The side lengths are AB = 3x + 1, AC = 5x − 11, and BC = 14. What is the perimeter?

Show worked solutionHide worked solution

Recognize the structure. The sides opposite B and C are AC and AB, respectively.

Work it through. Set their lengths equal:

$3x + 1 = 5x - 11 x = 6.$

Thus $AB = AC = 19,$ and the perimeter is $19 + 19 + 14 = 52.$ The side lengths satisfy the triangle inequality.

Answer: 52 units.

Key takeaway. The diagram-to-equation step is the real geometry. Once $AB = AC$ is justified, the remaining task is algebra and substitution.

2.04. Counting possible integer side lengths

Two sides of a triangle have lengths 7 and 12. How many integer values are possible for the third side length?

Show worked solutionHide worked solution

Recognize the structure. Apply both the upper and lower triangle-inequality bounds.

Work it through. If the third side is c, then

$|12 - 7| < c < 12 + 7 5 < c < 19.$

The integers are 6, 7, . . . , 18, a total of $18 - 6 + 1 = 13.$

Answer: 13.

Avoid the trap. Neither 5 nor 19 is allowed. At either endpoint the triangle would be flat. Also distinguish the largest possible integer, 18, from the number of possible integers.

2.05. Read congruence in the stated order

△ABC ≅ △FDE. If AB = 7, BC = 11, and AC = 9, what is the length of FE?

Show worked solutionHide worked solution

Recognize the structure. The order of the letters fixes the correspondence.

Work it through. Match A ↔ F, B ↔ D, and C ↔ E. Therefore FE corresponds to AC, not to AB. Congruent corresponding sides have equal lengths, so $FE = 9.$

Answer: 9 units.

Key takeaway. Writing the two vertex names in aligned rows is often faster and safer than trying to mentally rotate a figure. Congruence preserves correspondence through reflections as well as rotations.

2.06. Find the missing angle before scaling

In △ABC, ∠A = 41° and ∠B = 73°. In △DEF, ∠E = 73° and ∠F = 66°. If AB = 8, DE = 12, and BC = 10, find EF.

Show worked solutionHide worked solution

Recognize the structure. Complete the angle information to establish similarity and correspondence.

Work it through. In the first triangle, $∠C = 180 - 41 - 73 = 66°.$ In the second, $∠D = 41°.$ Thus △ABC ∼ △DEF, with scale factor $12/8 = 3/2$ from the first to the second. Since BC corresponds to EF,

$\frac{3}{2}$

Answer: 15 units.

Avoid the trap. An equal-angle pair identifies vertices, not necessarily the sides drawn in matching positions on the page.

2.07. The missing condition for SAS similarity

In two triangles, AB = 6, AC = 9, DE = 10, and DF = 15. Which additional statement is sufficient to prove △ABC ∼ △DEF by SAS similarity? A) ∠A = ∠D B) ∠C = ∠F C) BC = EF D) ∠A + ∠D = 180°

Show worked solutionHide worked solution

Recognize the structure. The known side pairs are proportional; SAS also requires the angle between them.

Work it through. The ratios are $AB/DE = 6/10 = 3/5$ and $AC/DF = 9/15 = 3/5.$ The included angles are ∠A and ∠D, because each lies between its two given sides. Their equality completes SAS similarity.

Answer: A.

Avoid the trap. A nonincluded angle, such as C or F, does not supply the SAS condition. Equal side ratios alone do not fix the angle between those sides.

2.08. Algebra inside a similarity proportion

A triangle with side lengths 6, 9, and 12 is similar to a second triangle. The sides corresponding to 6 and 9 have lengths x + 2 and 2x − 2, respectively. What is the second triangle’s perimeter?

Show worked solutionHide worked solution

Recognize the structure. Both corresponding side pairs must produce the same scale factor.

Work it through. Set up

$\frac{x + 2}{6}$

The scale factor is $(10 + 2)/6 = 2.$ Therefore the second perimeter is $2(6 + 9 + 12) = 54.$

Answer: 54 units.

Another efficient route. Find the three second-triangle sides directly: 12, 18, and 24. Their sum also gives 54.

2.09. Part of a side versus the whole side

In △ABC, D lies on AB, E lies on AC, and DE ∥ BC. If AD = 6, DB = 9, and AE = 8, what is EC?

Show worked solutionHide worked solution

Recognize the structure. The small triangle is ADE; its corresponding large triangle is ABC.

Work it through. The whole side AB is $6 + 9 = 15.$ Therefore

$\frac{AE}{AC}$

Since $AE = 8,$ $AC = 8(5/2) = 20.$ The requested remaining segment is $EC = 20 - 8 = 12.$

Answer: 12 units.

Another efficient route. The valid part-to-part proportion is $6/9 = 8/EC,$ which also yields $EC = 12.$

Avoid the trap. The equation $6/9 = 8/AC$ mixes part-to-part and part-to-whole ratios.

2.10. A shadow model with explicit assumptions

On level ground, a vertical 1.8-meter pole casts a 2.4-meter shadow. At the same time and location, a vertical tree casts a 16-meter shadow. Assuming the sun’s rays are parallel, how tall is the tree?

Show worked solutionHide worked solution

Recognize the structure. Vertical objects and level ground create right triangles; parallel sunlight makes the other angles match.

Work it through. The two triangles are similar. Their height-to-shadow ratios are equal:

$\frac{h}{16}$

Answer: 12 meters.

Key takeaway. The modeling assumptions matter. Different slopes of ground or objects tilted away from vertical would not automatically produce these similar triangles.

2.11. Perimeters scale like side lengths

Two similar triangles have perimeters 42 and 70. A side of the smaller triangle has length 9. What is the corresponding side length in the larger triangle?

Show worked solutionHide worked solution

Recognize the structure. A perimeter is a sum of lengths, so it uses the same scale factor as each side.

Work it through. The scale factor from smaller to larger is $70/42 = 5/3.$ Thus the corresponding length is $9(5/3) = 15.$

Answer: 15 units.

Avoid the trap. Do not take a square root of a perimeter ratio. Square roots convert area ratios to length ratios.

2.12. Convert an area ratio to a height ratio

Two similar triangles have areas in the ratio 49 : 81, smaller to larger. A corresponding altitude of the larger triangle is 18. Find the smaller altitude.

Show worked solutionHide worked solution

Recognize the structure. Altitudes are lengths; first take the square root of the area ratio.

Work it through. The smaller-to-larger length ratio is

$\sqrt{\frac{49}{81}}$

The smaller altitude is therefore $18(7/9) = 14.$

Answer: 14 units.

Key takeaway. The same 7/9 ratio applies to corresponding sides and perimeters. The 49/81 ratio applies only to areas.

2.13. An altitude reveals three similar triangles

In right triangle ABC, ∠C = 90°. Altitude CD meets hypotenuse AB at D, with AD = 9 and DB = 16. Find CD.

Show worked solutionHide worked solution

Recognize the structure. Each smaller right triangle shares an acute angle with the original triangle, so AA similarity applies.

Work it through. Matching the two small triangles gives $AD/CD = CD/DB.$ With $CD = h,$

$\frac{9}{h}$

Choose the positive root because h is a length.

Answer: 12 units.

Key takeaway. The useful shortcut $h 2 = (AD)(DB)$ comes from similarity. It applies to an altitude to the hypotenuse of a right triangle, not an arbitrary triangle altitude.

2.14. Sufficient evidence, not a suggestive picture

Suppose AB = DE and AC = DF. Which additional statement guarantees that △ABC ≅ △DEF? A) ∠B = ∠E B) ∠A = ∠D C) Both triangles are obtuse D) Both figures have the same orientation

Show worked solutionHide worked solution

Recognize the structure. Two fixed sides become a fixed triangle when their included angle is fixed.

Work it through. The included angles are A and D. Their equality establishes SAS congruence, so B is sufficient.

Answer: B.

Avoid the trap. Choice A gives SSA, not SAS. This can be ambiguous: with a fixed side from B to A and a fixed ray from B, a circle centered at A can intersect that ray twice, giving two possible locations for C with the same AC. Neither obtuseness nor drawing orientation fixes the missing information.

Key takeaway. For a “guarantees” question, a condition must work in every allowed case, not merely in the sketch you imagined.

2.15. A trapezoid is the difference of similar triangles

In △ABC, points D and E lie on AB and AC, and DE ∥ BC. If AD/AB = 3/5 and the area of trapezoid DBCE is 64, what is the area of △ADE?

Show worked solutionHide worked solution

Recognize the structure. The given ratio is a length ratio; the trapezoid is the large triangle minus the small one.

Work it through. Let the large triangle’s area be T. The small triangle’s area is $(3/5) 2 T = 9T/25.$ Therefore

$\frac{9}{25}$

The small area is $9(100)/25 = 36.$

Answer: 36 square units.

Another efficient route. The small-triangle-to-trapezoid area ratio is 9 : 16, so the desired area is $64(9/16) = 36.$

Avoid the trap. Subtracting the length ratio 3/5 from 1 before squaring gives the wrong region ratio. Square first, then subtract.

Mastery check and error prevention

Before using a proportion, write a similarity statement or list the matching vertices. Before using k², state why the figures are similar. Before using a congruence rule, identify whether a given angle is included between the two given sides.

A reliable self-test is to rotate or reflect one triangle on your scratch paper. If your side matching changes even though the vertex names do not, you were matching by appearance instead of by correspondence.