Digital SAT Math · Geometry & Trigonometry

Triangles

Most triangle items on the Digital SAT are a proportion in disguise. Once two triangles share enough angles to be similar, the whole job is writing the similarity statement with vertices in matching order and reading the side proportion off that statement. Angle sum, exterior angles, and isosceles base angles are the tools that get you the equal angles; the payoff is almost always a missing length or a missing degree measure.

On the test


Domain	Geometry and Trigonometry (score report)
CB skill	Lines, angles, and triangles (triangle half)
What it looks like	A labeled triangle, two triangles that share an angle, or a small triangle nested inside a larger one by a line parallel to a side
Often asked	“What is the length of…?”, “What is the measure of angle…?”, “Which of the following could be the length of…?”
Format	Multiple choice and student-produced response
Calculator	Allowed; rarely needed once the proportion is set. Desmos can cross-multiply, but it will happily solve a proportion with the wrong sides paired
Figures	Most similarity and exterior-angle items ship a diagram; pure angle-sum and inequality items are often text-only

Recognition cues: similar, corresponding, parallel to, an exterior ray drawn past a vertex, tick marks for equal sides, could be the length, nested triangles sharing a vertex.

What this page owns

This page owns triangle-local work: angle sum $180^\circ$ , the exterior-angle theorem, isosceles base angles, triangle inequality, and using similarity to write a proportion once angles match. Formal similarity/congruence criteria as named theorems (SSS, SAS, HL) and area scale factors go to similarity-and-congruence. Right-triangle side ratios (SOH-CAH-TOA) and pure Pythagorean length hunts live on their own skills. Parallel-line transversal chase without a triangle sits on lines-and-angles.

Pattern recognition

Five shapes cover nearly every item:

Angle sum — two interior angles given; the third is $180^\circ$ minus their sum.
Exterior angle — an exterior angle equals the sum of the two remote (non-adjacent) interiors; it is also supplementary to the adjacent interior.
Isosceles base angles — equal sides force equal base angles (and the reverse); never from appearance alone.
Similar triangles / proportion — shared or equal angles → write $\triangle ABC \sim \triangle DEF$ in vertex order → pair corresponding sides → solve.
Triangle inequality — for sides $a, b, c$ , each is strictly less than the sum of the other two (equality is not a triangle).

Method

Mark every equal angle you can justify. Shared angles, vertical angles, angles from parallel lines, base angles from equal sides, and supplements of a linear pair are all fair game. Do not invent equal sides from how the figure looks.
If two triangles are similar, write the similarity statement with vertices in matching order. Same-position letters hold equal angles: $A \leftrightarrow D$ , $B \leftrightarrow E$ , $C \leftrightarrow F$ .
Build the proportion from that statement only. Corresponding sides are the segments joining corresponding vertices.
Solve, then check plausibility. A side of the smaller triangle cannot come out longer than its correspondent on the larger one. An interior angle must sit between $0^\circ$ and $180^\circ$ , and the three interiors must sum to $180^\circ$ .
On exterior-angle items, name the adjacent interior first. The exterior equals $180^\circ$ minus that adjacent angle, and also equals the sum of the other two interiors — use whichever is faster, and never treat the exterior as equal to one remote alone.
On “which length is possible,” test the strict inequalities. The longest side must be shorter than the sum of the other two; a candidate equal to that sum is a straight line, not a triangle.

Words / marks in the stem	What they force
similar / $\sim$	equal corresponding angles; proportional corresponding sides
tick marks on two sides	those sides are equal → base angles equal
a ray extending one side	exterior angle; equals sum of remotes
a line parallel to one side of a triangle	creates a small triangle similar to the large one
could be / possible for a third side	triangle inequality, strict
right-angle mark	$90^\circ$ at that vertex; may unlock nested similarity on the altitude figure

Worked example 1 — angle sum, then exterior

Stem. In $\triangle ABC$ , $m\angle A = 48^\circ$ and $m\angle B = 75^\circ$ . (a) What is $m\angle C$ ? (b) Side $BC$ is extended past $C$ to a point $D$ . What is $m\angle ACD$ ?

Step 1 — angle sum for the interior.

m\angle C = 180^\circ - 48^\circ - 75^\circ = 57^\circ

Step 2 — exterior as supplement of the adjacent interior.

m\angle ACD = 180^\circ - 57^\circ = 123^\circ

Step 3 — exterior as sum of remotes (check).

m\angle ACD = m\angle A + m\angle B = 48^\circ + 75^\circ = 123^\circ

Same number both ways.

Check. Interiors $48 + 75 + 57 = 180$ ✓. Exterior larger than each remote ✓. Answers: (a) $57^\circ$ ; (b) $123^\circ$ .

Trap watch. Reporting $123^\circ$ for part (a) answers the exterior. Reporting $57^\circ$ for part (b) answers the adjacent interior. Reporting $48^\circ + 57^\circ = 105^\circ$ as the exterior drops remote $B$ and includes the adjacent angle.

Worked example 2 — isosceles, then similar proportion

Stem. In $\triangle ABC$ , $AB = AC$ and $m\angle A = 50^\circ$ . A second triangle $\triangle DEF$ is similar to $\triangle ABC$ with correspondence $A \leftrightarrow D$ , $B \leftrightarrow E$ , $C \leftrightarrow F$ . If $AB = 8$ and $DE = 12$ , and $BC = 10$ , what is $EF$ ?

Step 1 — base angles of the isosceles triangle. Equal sides $AB = AC$ force equal base angles:

m\angle B = m\angle C = \frac{180^\circ - 50^\circ}{2} = 65^\circ.

(Useful if an angle were missing; here it confirms the shape before the scale.)

Step 2 — similarity statement and scale. $\triangle ABC \sim \triangle DEF$ , so

\frac{DE}{AB} = \frac{12}{8} = \frac{3}{2}.

Step 3 — corresponding sides. $BC$ joins $B,C$ and pairs with $EF$ joining $E,F$ :

EF = BC \cdot \frac{3}{2} = 10 \cdot \frac{3}{2} = 15.

Check. Larger triangle has longer sides; $15 > 10$ ✓. Pairing $BC$ with $DE$ would give $10 \cdot 8/12 \approx 6.67$ , which shrinks the side that should grow — a correspondence error. Answer: $15$ .

Trap watch. $10 \cdot 8/12$ inverts the scale. $12$ is $DE$ , not $EF$ . $65$ is a base angle from Step 1, not a length.

Worked example 3 — nested similar from a parallel

Stem. In $\triangle CAB$ , points $D$ and $E$ lie on sides $CA$ and $CB$ respectively, and $DE \parallel AB$ . Given $CD = 4$ , $DA = 6$ , and $DE = 6$ , what is $AB$ ?

Step 1 — name the similar pair. Shared $\angle C$ , and corresponding angles equal because $DE \parallel AB$ , so

\triangle CDE \sim \triangle CAB

with vertices in that order ( $C \leftrightarrow C$ , $D \leftrightarrow A$ , $E \leftrightarrow B$ ).

Step 2 — full side of the large triangle. $CA = CD + DA = 4 + 6 = 10$ . The ratio uses $CA$ , not $DA$ :

\frac{CD}{CA} = \frac{4}{10} = \frac{2}{5}.

Step 3 — proportion for the bases.

\frac{DE}{AB} = \frac{2}{5} \quad\Rightarrow\quad \frac{6}{AB} = \frac{2}{5} \quad\Rightarrow\quad AB = 15.

Check. Small base $6$ is $\frac{2}{5}$ of $15$ ✓. Using $DA$ instead of $CA$ would give ratio $4/6$ and $AB = 9$ , which under-scales. Answer: $15$ .

Trap watch. $AB = 6$ treats the triangles as congruent. $AB = 6 \cdot 4/10 = 2.4$ inverts the ratio. $AB = 6 + 6 = 12$ adds instead of scaling.

Practice

Answer before opening the explanation. Two items are student-produced response. Eight items ship a triangle diagram — nested similar pairs, an exterior ray, isosceles marks, or a right triangle with an altitude — because on this skill the figure usually is the question. Every wrong choice below is a named trap: sides out of correspondence, exterior misapplied, isosceles from appearance, or an inequality boundary treated as allowed. Log the trap name when you miss one.

12 questions — 10 multiple choice, 2 student-produced response. Every wrong choice has its own explanation.

Question 1 Warm-up

In triangle ABC, the measure of angle A is 48° and the measure of angle B is 75°. What is the measure of angle C, in degrees?

Show the answer Choice A

Why it is right

The interior angles of any triangle sum to 180°. So m∠C = 180 − 48 − 75 = 57. Check: 48 + 75 + 57 = 180. The answer is an interior angle strictly between 0° and 180°, and it is smaller than the exterior that would form by extending a side past C (that exterior would be 48 + 75 = 123).

Why each other choice fails

Choice B: Copies angle A. The two given angles are inputs to the sum, not answers unless the triangle is isosceles with those equal — and nothing in the stem says that.
Choice C: Copies angle B. Same error as B: a given angle reported as the missing one.
Choice D: Computes the exterior angle at C (sum of the remotes 48 + 75) instead of the interior. The interior is 180 − 123 = 57, not 123.

Question 2 Standard

In the figure, side AB of triangle ABC is extended to point D. The measure of angle A is 42° and the measure of angle C is 67°. What is the measure of exterior angle CBD, in degrees?

Figure. Triangle ABC with exterior angle at B formed by extending AB to D.

Show the answer Choice B

Why it is right

An exterior angle of a triangle equals the sum of the two remote interior angles. The remotes to exterior ∠CBD are ∠A and ∠C, so m∠CBD = 42 + 67 = 109. Check via the adjacent interior at B: m∠ABC = 180 − 42 − 67 = 71, and the exterior is supplementary to that adjacent interior, 180 − 71 = 109. Both routes agree.

Why each other choice fails

Choice A: Reports the adjacent interior angle at B (180 − 42 − 67 = 71) instead of the exterior. The exterior is the supplement of 71, not 71 itself.
Choice C: Subtracts the given angles (67 − 42 = 25). An exterior angle is a sum of remotes, not a difference.
Choice D: Adds one remote to the adjacent interior idea incorrectly, or computes 180 − 42 = 138 — supplement of only one angle, not the exterior theorem.

Question 3 Standard

In the figure, triangle ABC is isosceles with AB = AC, and the measure of angle A is 50°. What is the measure of angle B, in degrees?

Figure. Isosceles triangle ABC with AB = AC and vertex angle 50°.

Show the answer Choice C

Why it is right

Equal sides AB = AC force equal base angles at B and C. Those two base angles share the remaining 180 − 50 = 130°, so each is 130 / 2 = 65°. Check: 50 + 65 + 65 = 180. The figure’s tick marks (not appearance alone) justify the equal sides.

Why each other choice fails

Choice A: Assumes the triangle is equilateral, or copies the vertex angle as a base angle. Equal sides make the base angles equal to each other, not necessarily equal to the vertex.
Choice B: Stops after subtracting the vertex from 180 (130°) and reports the sum of the two base angles instead of one base angle.
Choice D: Doubles the vertex angle (2 × 50). There is no rule that a base angle is twice the vertex; that would also break the 180° sum (50 + 100 + 100 = 250).

Question 4 Standard

In the figure, triangle ABC is similar to triangle DEF with correspondence A → D, B → E, and C → F. The length of AB is 8, the length of BC is 10, and the length of DE is 12. What is the length of EF?

Figure. Similar triangles ABC and DEF with AB = 8, BC = 10, and DE = 12.

Show the answer Choice D

Why it is right

Similarity △ABC ∼ △DEF pairs AB with DE and BC with EF. The scale factor from ABC to DEF is DE/AB = 12/8 = 3/2. Therefore EF = BC × 3/2 = 10 × 3/2 = 15. Plausibility: the image triangle is larger (scale 1.5), so EF must exceed BC.

Why each other choice fails

Choice A: Inverts the scale: multiplies 10 by 8/12 to get 20/3 ≈ 6.67. That shrinks the side that should grow under a scale factor greater than 1.
Choice B: Copies DE. DE corresponds to AB, not to BC; EF is the image of BC.
Choice C: Copies BC without scaling. Correspondence requires multiplying by 12/8.

Question 5 Standard

Two sides of a triangle have lengths 7 and 12. Which of the following could be the length of the third side?

Show the answer Choice A

Why it is right

For sides 7, 12, and x to form a triangle, the strict inequalities |12 − 7| < x < 12 + 7 must hold: 5 < x < 19. Among the choices only 12 lies strictly between 5 and 19. Check: 7 + 12 > 12, 7 + 12 > 12, and 12 + 12 > 7 all hold.

Why each other choice fails

Choice B: 4 is less than the difference 5, so 7 + 4 = 11 < 12 — the sides cannot close.
Choice C: 5 equals |12 − 7|, which makes a degenerate flat figure (7 + 5 = 12), not a triangle. The inequality is strict.
Choice D: 19 equals 12 + 7, another degenerate case (the sides lie on one line). Need x < 19.

Question 6 Harder

In the figure, points D and E lie on sides CA and CB of triangle CAB, respectively, and DE is parallel to AB. The length of CD is 4, the length of DA is 6, and the length of DE is 6. What is the length of AB?

Figure. Nested triangle CDE inside triangle CAB with DE parallel to AB.

Show the answer Choice B

Why it is right

Because DE ∥ AB, △CDE ∼ △CAB by AA (shared ∠C and corresponding angles with the parallel). Vertices match C→C, D→A, E→B. The full side CA is CD + DA = 10, so the ratio is CD/CA = 4/10 = 2/5. Then DE/AB = 2/5, so 6/AB = 2/5 and AB = 15. Check: 6 is exactly 2/5 of 15.

Why each other choice fails

Choice A: Uses DA in place of CA in the ratio (4/6), giving AB = 6 × 6/4 = 9. The large triangle’s corresponding side is the full CA = 10, not the leftover segment DA.
Choice C: Treats the triangles as congruent and copies DE. The parallel creates similarity with scale 2/5, not congruence.
Choice D: Inverts the proportion: AB = 6 × 4/10 = 2.4, making the large base shorter than the small base.

Question 7 Harder

In the figure, right triangle ABC with right angle at B is similar to right triangle DEF with right angle at E. The correspondence of vertices is A → D, B → E, and C → F. The legs of triangle ABC are AB = 6 and BC = 8, and the hypotenuse is AC = 10. If DE = 9, what is the length of EF?

Figure. Similar right triangles ABC and DEF with correspondence A→D, B→E, C→F.

Show the answer Choice C

Why it is right

Under △ABC ∼ △DEF with A→D, B→E, C→F, leg AB pairs with leg DE and leg BC pairs with leg EF. Scale factor is DE/AB = 9/6 = 3/2. So EF = BC × 3/2 = 8 × 3/2 = 12. Check: the hypotenuse scales to 10 × 3/2 = 15, and 9-12-15 is the 3-4-5 triple times 3, so the triple closes.

Why each other choice fails

Choice A: Scales the hypotenuse instead of leg BC: 10 × 9/6 = 15. That length is DF (image of AC), not EF.
Choice B: Builds a proportion with the legs swapped relative to the hypotenuse: 6 × 10/8 = 7.5. That ignores the given DE = 9 and pairs sides that the correspondence statement does not authorize.
Choice D: Applies the scale factor 3/2 to DE itself (9 × 3/2 = 13.5) instead of to BC. DE is already a side of the image triangle; the unknown is EF.

Question 8 Harder Student-produced response

Triangle RST is similar to triangle UVW with correspondence R → U, S → V, and T → W. The length of RS is 5, the length of ST is 8, and the length of UV is 20. What is the length of VW?

Your answer

Show the answer 32

Why it is right

Similarity △RST ∼ △UVW pairs RS with UV and ST with VW. Scale factor from RST to UVW is UV/RS = 20/5 = 4. Therefore VW = ST × 4 = 8 × 4 = 32. Check: 32/8 = 20/5 = 4, so the ratios match. The image is larger, and 32 > 8 as required.

Answers students type instead

2: Inverts the scale: 8 × 5/20 = 2. That answers a shrink when the given UV = 20 is four times RS = 5.
20: Copies UV. UV corresponds to RS, not to ST; VW is the image of ST.
12.5: Pairs ST with UV incorrectly: 8 × 20/ 12.8 or 5 × 20/8 = 12.5 — uses RS as if it corresponded to ST when cross-multiplying.

Question 9 Harder

In the figure, triangle ABC is isosceles with AB = AC. Side BC is extended to point D, and the measure of exterior angle ACD is 110°. What is the measure of angle BAC, in degrees?

Figure. Isosceles triangle ABC (AB = AC) with exterior angle 110° at C.

Show the answer Choice D

Why it is right

Exterior ∠ACD and interior ∠ACB form a linear pair, so m∠ACB = 180 − 110 = 70°. Because AB = AC, base angles are equal: m∠ABC = m∠ACB = 70°. Then m∠BAC = 180 − 70 − 70 = 40°. Check via remotes: exterior 110° = ∠BAC + ∠ABC = y + 70, so y = 40. Both routes agree.

Why each other choice fails

Choice A: Copies the exterior measure as the vertex angle. The exterior is at C, not at A.
Choice B: Reports a base angle (70°) instead of the vertex. After finding the base angles, the question still asks for ∠BAC.
Choice C: Halves the exterior (110/2 = 55). The exterior equals the sum of the two remotes, not twice the vertex alone unless the triangle is isosceles in a different way — here remotes are vertex + one base angle.

Question 10 Hardest

In the figure, points F and G lie on sides AB and AC of triangle ABC, respectively, and FG is parallel to BC. The length of AF is 5, the length of FB is 7, and the length of FG is 10. What is the length of BC?

Figure. Nested triangle AFG inside triangle ABC with FG parallel to BC.

Show the answer Choice A

Why it is right

FG ∥ BC implies △AFG ∼ △ABC with A→A, F→B, G→C. Full side AB = AF + FB = 5 + 7 = 12, so the ratio is AF/AB = 5/12. Then FG/BC = 5/12, so 10/BC = 5/12 and BC = 10 × 12/5 = 24. Check: 10 is 5/12 of 24. Using only FB = 7 in the denominator is the standard nested-ratio error.

Why each other choice fails

Choice B: Uses FB instead of AB: 10 × 7/5 = 14. The similarity ratio is AF to the full AB = 12, not AF to the leftover segment.
Choice C: Uses AB ≈ 12 but inverts or mis-divides: 10 × 12/7 ≈ 17.1, putting FB in the denominator as if the ratio were AF/FB.
Choice D: Adds the three given numbers or treats FG as half of something (10 + 7 − 5) / wrong scale — 12 is AB itself, the length along the side, not the base BC.

Question 11 Hardest Student-produced response

In triangle CAB, points D and E lie on sides CA and CB, respectively, and DE is parallel to AB. The length of CD is 6, the length of DA is 4, and the length of CE is 9. What is the length of CB?

Your answer

Show the answer 15

Why it is right

DE ∥ AB gives △CDE ∼ △CAB with ratio CD/CA. CA = CD + DA = 6 + 4 = 10, so CD/CA = 6/10 = 3/5. Corresponding sides CE and CB are in the same ratio: CE/CB = 3/5, so 9/CB = 3/5 and CB = 9 × 5/3 = 15. Check: 9/15 = 3/5, matching 6/10.

Answers students type instead

6: Inverts with the short segment: 9 × 4/6 = 6, making CB shorter than CE.
13.5: Uses DA in the ratio in place of CA: 9 × 6/4 = 13.5. The large side is the full CA = 10.
22.5: Uses DA as the large side incorrectly the other way: 9 × 10/4 = 22.5 — mixes full CA in the numerator path with DA in the denominator.

Question 12 Hardest

In the figure, triangle ABC is a right triangle with right angle at C. The legs have lengths AC = 15 and BC = 20, and the hypotenuse AB has length 25. The altitude from C to AB meets AB at D. What is the length of AD?

Figure. Right triangle ABC with altitude from C to hypotenuse AB.

Show the answer Choice B

Why it is right

The altitude to the hypotenuse creates three similar triangles: △ACD ∼ △ABC ∼ △CBD. From △ACD ∼ △ABC with A→A, C→B, D→C is one ordering; the useful geometric-mean relation is AC² = AD · AB. So 15² = AD · 25, 225 = 25 · AD, and AD = 9. Check: DB = AB − AD = 16, and BC² = DB · AB → 400 = 16 · 25, true. Also CD² = AD · DB = 9 · 16 = 144.

Why each other choice fails

Choice A: Reports the altitude CD (which is 12) instead of AD. CD is the geometric mean of the two hypotenuse segments, not the segment adjacent to A.
Choice C: Reports DB = 16 (from BC² = DB · AB → DB = 400/25 = 16) — the other hypotenuse segment, not AD.
Choice D: Copies leg AC = 15. Similarity relates AC to the hypotenuse and to AD; AC is not equal to AD.

Common mistakes

Sides paired out of correspondence order — a correct scale factor applied to the wrong side pair because the similarity statement was never written (or was written with letters scrambled).
Exterior angle treated as the adjacent interior — reporting $180^\circ - \text{exterior}$ when the question asked for the exterior, or the reverse.
Exterior built from the wrong two angles — adding the adjacent interior into the remote sum, or using only one remote.
Isosceles assumed from appearance — equal base angles (or equal sides) claimed because the drawing “looks” symmetric, with no tick marks or stated equal sides.
Triangle inequality with equality allowed — accepting a third side equal to the sum (or difference) of the other two; that is a line segment, not a triangle.
Nested ratio using the wrong segment — $DA$ in place of $CA$ , or midsegment instincts when the parallel is not at half height.
Inverted proportion — large over small on one side of the equation and small over large on the other.
Reporting an intermediate angle as the answer — a base angle when the vertex was asked, or a remote when the exterior was asked.
Additive scaling — adding the same amount to a side that should be multiplied by a scale factor.
Ignoring the picture’s size check — a “solution” that makes the small triangle’s side longer than the large triangle’s corresponding side.

FAQ

Do I need to memorize SSS, SAS, and AA by name? You need AA as a working habit: two pairs of equal angles force similarity. Named congruence criteria and full similarity proofs are the neighbouring skill; here, once angles match, write the vertex-order statement and proportion.

When is an exterior angle equal to 180° minus something, and when is it a sum? Always both. It is supplementary to the adjacent interior and equal to the sum of the two remotes. Pick the route with the numbers you have.

Can three lengths 3, 4, 7 form a triangle? No. $3 + 4 = 7$ , so the sides lie on a single line. You need $3 + 4 > 7$ .

How do I know which sides correspond? Only from the similarity statement. If $\triangle ABC \sim \triangle XYZ$ , then $AB/XY = BC/YZ = AC/XZ$ . Matching by “both are the bottom side in the picture” is how correspondence errors start.

Does a parallel line inside a triangle always create similar triangles? Yes — the small triangle that shares the apex with the large one is similar to the large one (AA). Midlines and midsegments are special cases where the ratio is $\frac12$ ; do not assume $\frac12$ unless the parallel is halfway or the stem says so.

Grid-in format? Enter the exact length or degree measure. The expensive SPR error on this skill is a correspondence flip that produces a clean wrong number with no lettered choices to warn you.

Editorial team · last reviewed 2026-08-12

Triangles

On the test

What this page owns

Pattern recognition

Method

Worked example 1 — angle sum, then exterior

Worked example 2 — isosceles, then similar proportion

Worked example 3 — nested similar from a parallel

Practice

Question 1 Warm-up

Why it is right

Why each other choice fails

Question 2 Standard

Why it is right

Why each other choice fails

Question 3 Standard

Why it is right

Why each other choice fails

Question 4 Standard

Why it is right

Why each other choice fails

Question 5 Standard

Why it is right

Why each other choice fails

Question 6 Harder

Why it is right

Why each other choice fails

Question 7 Harder

Why it is right

Why each other choice fails

Question 8 Harder Student-produced response

Why it is right

Answers students type instead

Question 9 Harder

Why it is right

Why each other choice fails

Question 10 Hardest

Why it is right

Why each other choice fails

Question 11 Hardest Student-produced response

Why it is right

Answers students type instead

Question 12 Hardest

Why it is right

Why each other choice fails

Common mistakes

FAQ

Related skills