Question 16·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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For triangles $ABC$ and $DEF$ , angles $B$ and $E$ each measure $48^\circ$ , the length of $\overline{AB}$ is $14$ , and the length of $\overline{DE}$ is $21$ . Which of the following additional pieces of information are sufficient to prove that triangle $ABC$ is similar to triangle $DEF$ ?

I. The length of $\overline{BC}$ is $16$ , and the length of $\overline{EF}$ is $24$ .

II. The length of $\overline{AC}$ is $12$ , and the length of $\overline{DF}$ is $18$ .

III. An exterior angle at $A$ has the same measure as an exterior angle at $D$ .

For triangle similarity, start by matching the vertices and sides, then test each added fact together with what you already know. Use Desmos to compare side ratios. Two proportional side pairs and an equal angle prove similarity by SAS only when the angle is between those sides; a second pair of equal angles proves it by AA.

Hints

Hint 1
The triangle names give the corresponding vertices in order: $A$ with $D$ , $B$ with $E$ , and $C$ with $F$ . Which sides meet at the given equal angles?
Hint 2
For statement II, don't stop when the side ratios match. Check whether the $48^\circ$ angle is between the two sides whose lengths you know in each triangle.
Hint 3
An exterior angle and the interior angle beside it add to $180^\circ$ . If the exterior angles at $A$ and $D$ match, what does that tell you about the interior angles at those vertices?

Step-by-step

Test SAS, SSA, and AA
Step 1
Match the parts of the triangles
To prove similarity in the order $ABC$ and $DEF$ , match $A$ with $D$ , $B$ with $E$ , and $C$ with $F$ . So the corresponding sides are $AB$ with $DE$ , $BC$ with $EF$ , and $AC$ with $DF$ . The given equal angles are at $B$ and $E$ .
Step 2
Compare the side ratios in I
For I, type $\frac{21}{14}$ and $\frac{24}{16}$ in Desmos. It shows $1.5$ for each. So $DE$ is $1.5$ times $AB$ , and $EF$ is $1.5$ times $BC$ : the two matching side pairs have the same scale factor.
Step 3
Use the angle between those sides
Sides $AB$ and $BC$ meet at $B$ ; $DE$ and $EF$ meet at $E$ . Those are the equal $48^\circ$ angles, so each angle lies between the proportional sides. This is side-angle-side (SAS) similarity, making I sufficient.
Step 4
Check the remaining side ratio
For II, add $\frac{18}{12}$ in Desmos. It also shows $1.5$ , so $DF$ and $AC$ have the same scale factor as $DE$ and $AB$ . The matching ratio is real, but you still need to check where the equal angle sits.
Step 5
See why II does not fix the shape
The angle at $B$ is not between $AB$ and $AC$ : those sides meet at $A$ . This is side-side-angle (SSA), not SAS. With these lengths, $C$ can occupy two positions along the ray from $B$ while $AB=14$ , $AC=12$ , and angle $B=48^\circ$ stay fixed. Proportional sides and an angle across from one of them do not guarantee similarity. So II is insufficient.
Step 6
Turn III into an interior-angle match
Let $e$ be the shared measure of the exterior angles in III. An exterior angle and the interior angle beside it make a straight angle, so $m\angle A=180^\circ-e=m\angle D$ . The interior angles at $A$ and $D$ match.
Step 7
Finish with two angle pairs
Angles $B$ and $E$ already match, and III makes angles $A$ and $D$ match. Angle-angle (AA) similarity says two matching angle pairs are enough, so III is sufficient. I and III are sufficient, but II is not. Choice D.

Question 16·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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