Question 9·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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In equilateral triangle $ABC$ , point $D$ lies on segment $BC$ such that

\frac{BD}{BC}=\frac12.

Segment $AD$ is drawn. Which choice gives the measure of angle $DAB$ ?

An angle bisector splits an angle into two equal parts. In an equilateral triangle, a segment from the top vertex to the midpoint of the base may be one. Use the fraction to locate the midpoint, then show the two smaller triangles match by three equal sides. Halving a side alone doesn't always halve an angle.

Hints

Hint 1
A midpoint splits a segment into two equal pieces. The fraction says $BD$ is half the base. Since $D$ lies between $B$ and $C$ , how does $DC$ compare with $BD$ ?
Hint 2
Three matching pairs of sides make triangles congruent, meaning the same size and shape. Compare $AB$ with $AC$ , $BD$ with $DC$ , and the shared $AD$ . What does that tell you about the two angles at $A$ ?

Step-by-step

Match the two smaller triangles
Step 1
Show that D is the midpoint
No Desmos needed. Matching triangles give this angle directly. The given fraction means $BD$ is half of $BC$ :
$BD=\frac{BC}{2}.$
Because $D$ lies on $BC$ , the remaining piece $DC$ is the other half. So $BD=DC$ , and $D$ is the midpoint of $BC$ .
Step 2
Prove that AD splits the top angle equally
Equilateral means $AB=AC$ . Triangles $ABD$ and $ACD$ also have $BD=DC$ and share $AD$ . By side-side-side congruence, three matching pairs of sides make the triangles the same size and shape, so their angles at $A$ match:
$m\angle DAB=m\angle DAC.$
Equal angles at $A$ make $AD$ an angle bisector. The midpoint alone wouldn't prove this without the equal sloping sides.
Step 3
Halve the angle at A
Every angle of an equilateral triangle is $60^\circ$ . Since $AD$ splits that whole angle at $A$ into two equal parts:
$m\angle DAB=\frac{60^\circ}{2}=30^\circ.$
Angle $DAB$ measures $30^\circ$ . Choice A.

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Question 9·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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