Lines Angles and Triangles Pattern - Similar and Congruent Triangle Analysis

This pattern tests your ability to use similar triangles (same shape, different size) and congruent triangles (same shape and size) to find missing sides, angles, or ratios.

When Are Triangles Similar?

Two triangles are similar ($\triangle ABC \sim \triangle DEF$) if their corresponding angles are equal, which also means their corresponding sides are proportional. The main ways to prove similarity:

• AA (Angle-Angle): Two pairs of corresponding angles are equal (the third pair automatically matches since angles sum to $180°$)
• SSS Similarity: All three pairs of corresponding sides are in the same ratio
• SAS Similarity: Two pairs of sides are in the same ratio AND the angles between them are equal

On the SAT, AA similarity is by far the most common method.

Finding Angles in Similar Triangles

In similar triangles, corresponding angles are equal. The vertex order tells you the correspondence.

In the figure, right triangles $ABC$ and $DEF$ are similar, and vertex $A$ corresponds to vertex $D$. If the measure of angle $B$ is $35°$, what is the measure of angle $D$?

A) $35°$
B) $55°$
C) $65°$
D) $145°$

In $\triangle ABC$: $\angle A = 180° - 90° - 35° = 55°$.
Since $A$ corresponds to $D$: $\angle D = \angle A = 55°$.
The answer is B. Option A ($35°$) confuses $D$ with $B$'s correspondence. Always check the vertex order in the similarity statement.

Setting Up Proportions with Similar Triangles

When triangles are similar, corresponding sides are proportional. Write the ratio as $\dfrac{\text{side in small }\triangle}{\text{corresponding side in large }\triangle}$ and cross-multiply.

In the figure shown, segment $DE$ is parallel to segment $BC$, with point $D$ on side $AB$ and point $E$ on side $AC$. The length of $AD$ is $12$, the length of $DB$ is $8$, and the length of $DE$ is $18$. What is the length of segment $BC$?

A) $12$
B) $22.5$
C) $30$
D) $10.8$

Since $DE \parallel BC$, $\triangle ADE \sim \triangle ABC$ (by AA — corresponding angles are equal).
$AB = AD + DB = 12 + 8 = 20$
Set up the proportion of corresponding sides:
$\dfrac{AD}{AB} = \dfrac{DE}{BC}$
$\dfrac{12}{20} = \dfrac{18}{BC}$
$12 \cdot BC = 20 \cdot 18 = 360$
$BC = 30$
The answer is C.

Key rule: A line parallel to one side of a triangle cuts the other two sides proportionally. This is the Basic Proportionality Theorem (also called Thales' theorem on the SAT).

Parallel Lines and Proportional Segments

In the figure shown, $DE \parallel BC$. The length of $AD$ is $8$, the length of $DB$ is $12$, and the length of $EC$ is $15$. What is the length of segment $AC$?

A) $10$
B) $22$
C) $25$
D) $45$

$\triangle ADE \sim \triangle ABC$.
$AB = 8 + 12 = 20$.
Let $AE = x$. Then $AC = x + 15$.
$\dfrac{AD}{AB} = \dfrac{AE}{AC}$
$\dfrac{8}{20} = \dfrac{x}{x + 15}$
$8(x + 15) = 20x$
$8x + 120 = 20x$
$120 = 12x$
$x = 10$, so $AC = 10 + 15 = 25$.
The answer is C.

Gotcha: Make sure you're solving for the right length. If the question asks for $AC$ (the whole side), don't give $AE$ (just part of it).

Dilations Preserve Angles

A dilation changes size but not shape. Angles in the dilated figure are identical to the original. Side lengths scale by the given factor, but angles do not.

A company's logo features triangle $PQR$. A designer creates triangle $STU$ by dilating triangle $PQR$ by a scale factor of $2$. The vertices correspond as $P \to S$, $Q \to T$, $R \to U$. In the original, $\angle Q = 80°$ and $PR = 15$ cm. What is the measure, in degrees, of angle $T$?

A) $40°$
B) $80°$
C) $100°$
D) $160°$

Dilation preserves angles. $Q$ corresponds to $T$, so $\angle T = \angle Q = 80°$. The answer is B. The scale factor of $2$ and the side length $PR = 15$ are extra information — they affect sides, not angles.

Real-World Application: Shadow Problems

A classic SAT setup: a tall object and a shorter object both cast shadows from the same light source. The two triangles formed (light source, top of object, tip of shadow) are similar by AA.

A person who is $6$ feet tall stands on flat ground $12$ feet away from a $24$-foot-tall vertical lamppost. What is the length, in feet, of the shadow cast by the person?

A) $3$
B) $6$
C) $4$
D) $12$

The lamppost-and-shadow triangle is similar to the person-and-shadow triangle (both have right angles at the ground, and they share the angle at the shadow tip).
Let $x$ = shadow length.
$\dfrac{\text{lamppost height}}{\text{person height}} = \dfrac{\text{lamppost to shadow tip}}{\text{shadow length}}$
$\dfrac{24}{6} = \dfrac{12 + x}{x}$
$4x = 12 + x$
$3x = 12$
$x = 4$
The answer is C.

Setup tip: The total distance from the lamppost to the shadow tip is $12 + x$ (distance from lamppost to person, plus shadow length).

Hard: Altitude to the Hypotenuse

When you drop an altitude from the right angle to the hypotenuse in a right triangle, it creates two smaller triangles that are each similar to the original (and to each other). This unlocks powerful proportions.

In $\triangle PQR$, the measure of $\angle PQR$ is $90°$, and $QS$ is the altitude to the hypotenuse $PR$. The length of $PQ$ is $10$. The length of $PR$ is $15$ units greater than $PQ$. What is the value of the ratio $\dfrac{QR}{QS}$?

A) $\dfrac{2}{3}$
B) $\dfrac{3}{2}$
C) $\dfrac{5}{2}$
D) $\dfrac{2}{5}$

$PQ = 10$ and $PR = 10 + 15 = 25$.
By the Pythagorean theorem: $QR = \sqrt{PR^2 - PQ^2} = \sqrt{625 - 100} = \sqrt{525}$.
The altitude creates similar triangles: $\triangle QSR \sim \triangle PQR$ (they share $\angle R$ and both have a right angle).
From the similarity: $\dfrac{QR}{PR} = \dfrac{QS}{PQ}$, so $\dfrac{QS}{PQ} = \dfrac{QR}{PR}$.
Then $\dfrac{QR}{QS} = \dfrac{PR}{PQ} = \dfrac{25}{10} = \dfrac{5}{2}$.
The answer is C.

What to Do on Test Day

• Check the similarity statement order. In $\triangle ABC \sim \triangle DEF$: $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$. Corresponding sides are $AB/DE$, $BC/EF$, $AC/DF$
• Parallel line inside a triangle = similar triangles. If $DE \parallel BC$ in $\triangle ABC$, then $\triangle ADE \sim \triangle ABC$. Use the proportion $\dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{DE}{BC}$
• Set up proportions carefully. Write small triangle over big triangle (or vice versa) consistently. Cross-multiply to solve
• Angles don't scale. Dilations, similarity — angles stay the same. Only sides change. If they ask for an angle in a similar or dilated triangle, it equals the corresponding angle in the original
• Shadow/height problems: Draw the two right triangles. The heights are proportional to their base lengths. Watch whether the "base" includes the person's distance or not
• Altitude to hypotenuse in a right triangle creates three similar triangles. The key proportion is: each leg is the geometric mean of the hypotenuse and the adjacent hypotenuse segment
• Don't confuse "part" with "whole." If $AD = 8$ and $DB = 12$, then $AB = 20$. Use $AB$ in the proportion, not $DB$, when comparing to the full triangle