Use congruence and similarity criteria for triangles to solve problems and to justify relationships in figures that decompose into triangles

Learning goal: Use congruence and similarity criteria for triangles to solve problems and to justify relationships in geometric figures that can be decomposed into triangles.

The criteria, stated with their conditions. Two triangles are congruent (a sequence of rigid motions takes one onto the other) when SSS, three pairs of equal sides; SAS, two pairs of equal sides with the included angles equal; or ASA, two pairs of equal angles with the included sides equal, holds; AAS follows from ASA and the angle sum. Two triangles are similar when AA, two pairs of equal angles, holds; also when SSS for similarity, all three pairs of sides in one ratio, or SAS for similarity, two pairs of sides in one ratio with the included angles equal, holds. These two are established exactly as AA was: dilate the first triangle by the common ratio k, and the image is congruent to the second triangle by SSS or SAS congruence, so a dilation followed by rigid motions carries one triangle onto the other. Congruent triangles have all corresponding parts equal; similar triangles have all corresponding angles equal and all corresponding sides in one ratio. Two criteria that do not work: SSA (two sides and a non-included angle) does not guarantee congruence, and equal angles alone do not guarantee congruence, only similarity.

Method. Almost every problem in this lesson has the same shape. Find the triangles in the figure, or draw them in; the figure is decomposed into triangles. Identify a criterion that applies, by naming the equal angles or proportional sides with a reason for each. Write the correspondence of vertices. Then read off the fact you need, a length from a proportion or an equality of parts.

Problem type 1: indirect measurement. A person 1.8 m tall casts a shadow 2.4 m long at the same moment that a flagpole casts a shadow 9.6 m long on level ground. How tall is the flagpole? The person and shadow form a right triangle, and so do the flagpole and its shadow. The sun's rays are parallel, so the angle each shadow makes with the ray is the same, and both triangles have a right angle where the vertical meets the ground. By AA the triangles are similar. Corresponding sides: height/shadow is the same in both, so flagpole/9.6 = 1.8/2.4, and flagpole = 9.6 × 0.75 = 7.2 m. The condition "at the same moment" is what makes the sun angles equal; measured an hour apart, the triangles need not be similar.

Problem type 2: overlapping triangles that share vertical angles. Segments AE and BD cross at C, with AB parallel to DE. Given AC = 6, CE = 9, BC = 4 and AB = 5, find CD and DE. Angles ACB and ECD are vertical angles, so equal. Angles BAC and DEC are alternate interior angles for the transversal AE across the parallel lines AB and DE, so equal. By AA, triangle ABC ~ triangle EDC, with A ↔ E, B ↔ D, C ↔ C. The scale factor from ABC to EDC is EC/AC = 9/6 = 1.5. So DC = 1.5 × BC = 6 and DE = 1.5 × AB = 7.5. Reading the correspondence carefully is the whole skill: BC pairs with DC, not with EC.

Problem type 3: justifying a relationship in a figure. Claim: the diagonals of a parallelogram bisect each other. Let ABCD be a parallelogram with diagonals AC and BD meeting at E. The figure decomposes into four triangles; use ABE and CDE. AB = CD because opposite sides of a parallelogram are equal. Angle BAE = angle DCE, alternate interior angles for the transversal AC across the parallel sides AB and DC. Angle ABE = angle CDE, alternate interior angles for the transversal BD across the same parallel sides. So triangle ABE ≅ triangle CDE by ASA, with A ↔ C, B ↔ D, E ↔ E. Corresponding parts: AE = CE and BE = DE, which says E is the midpoint of both diagonals. Notice that the argument used a congruence criterion, not a similarity criterion, because it needed equal lengths, not proportional ones.

Problem type 4: the midsegment of a triangle. Let D be the midpoint of AB and E the midpoint of AC in triangle ABC. Claim: DE is parallel to BC and DE = ½ BC. Triangles ADE and ABC share angle A, and AD/AB = AE/AC = 1/2, so they are similar by SAS for similarity, with D ↔ B and E ↔ C. Then angle ADE = angle ABC, so DE and BC make equal corresponding angles with the transversal AB, which makes DE parallel to BC. And DE/BC = 1/2. The same conclusion also follows from the dilation from A with scale factor 2, which sends D to B and E to C, and takes the line DE to the parallel line BC with the segment doubled.

Choosing between congruence and similarity. Ask what the problem wants. Equal lengths or equal angles in two triangles that have the same size: congruence. A length in a triangle that is a scaled copy of another: similarity, with a proportion. In the parallelogram the two triangles have equal sides, so congruence; in the shadow problem the triangles differ in size, so similarity. A common error is claiming equal sides in triangles that are only similar. A proportion between congruent triangles is not wrong, since congruent triangles are similar with ratio 1; it is just a longer route to the equal lengths.

The misconception. Students are tempted to use SSA. Two triangles with sides 8 and 10 and a 40° angle opposite the side 8 can be two genuinely different triangles, so SSA proves nothing; a later lesson on the Law of Sines shows exactly the two possibilities. Before writing "congruent by SSA", find another pair of equal parts, or reconsider which angle is included between the sides.

Worked, shadow problem: A vertical stick 1.2 m tall casts a shadow 0.9 m long; at the same moment a building casts a shadow 15 m long on level ground. The stick and building are both perpendicular to the ground and the sun’s rays are parallel, so the two right triangles have equal angles at the tips of the shadows: similar by AA. height/shadow is constant: building/15 = 1.2/0.9 = 4/3, so the building is 20 m tall.

Worked, crossing segments: AE and BD cross at C, with AB parallel to DE, AC = 6, CE = 9, BC = 4, AB = 5. Vertical angles at C are equal; alternate interior angles BAC and DEC are equal. Triangle ABC ~ triangle EDC by AA with A ↔ E, B ↔ D. Scale factor 9/6 = 1.5. So CD = 1.5 × 4 = 6 and DE = 1.5 × 5 = 7.5. Check: AC/EC = 6/9, BC/DC = 4/6, AB/ED = 5/7.5, all 2/3.

Worked, diagonals of a parallelogram: ABCD with diagonals meeting at E. AB = CD (opposite sides), angle BAE = angle DCE and angle ABE = angle CDE (alternate interior angles across the parallel sides AB and DC). ASA gives triangle ABE ≅ triangle CDE, hence AE = CE and BE = DE: the diagonals bisect each other. If AC = 14, then AE = 7, without any further measurement.

Worked, a mirror on the ground: A student whose eyes are 1.5 m above the ground places a mirror on level ground 12 m from the base of a tree, then moves along the line through the tree and the mirror, away from the tree, until she sees the top of the tree in the mirror. At that moment her feet are 2 m from the mirror (hypothetical measurements). Light reflects at equal angles, so the angle at the mirror is the same on both sides, and both triangles have a right angle at the ground. By AA they are similar: tree/12 = 1.5/2, so the tree is 9 m tall.

Recording task: Draw segments AE and BD crossing at C with AB parallel to DE, using AC = 6 cm, CE = 9 cm, BC = 4 cm and AB = 5 cm. Measure CD and DE and record them beside the predicted 6 cm and 7.5 cm. Record a two-column table listing each pair of equal angles with its reason (vertical angles; alternate interior angles) and the similarity statement with the vertices in corresponding order. Then draw a parallelogram, its two diagonals, and measure the four half-diagonals; record that the two halves of each diagonal agree, and write the ASA justification in three lines.

Independent performance: Solve three problems and justify one relationship. Problems: a 1.8 m person with a 2.4 m shadow beside a pole with a 9.6 m shadow; two triangles formed by crossing segments with a pair of parallel sides and three known lengths of your choice; and a triangle with a midsegment where the third side is 13 cm. Relationship: prove that the diagonals of a parallelogram bisect each other, or that the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length. Success criteria: each problem names the criterion (AA, SAS for similarity, ASA) with a reason for every equal angle or proportional pair; the correspondence of vertices is written before any proportion; each length is checked by a second ratio; the proof states which figure is decomposed into which triangles; and one sentence explains why SSA was not used.