Super-Hard SAT Math Question 152 | Free SAT Prep | Aniko

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Question 152·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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In the figure, lines $a$ and $b$ are parallel. Points $A$ and $D$ lie on line $a$ , and points $B$ and $C$ lie on line $b$ . Segments $\overline{AB}$ , $\overline{BC}$ , $\overline{CD}$ , and $\overline{DA}$ form a trapezoid, and diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E$ .

Given that $\angle ABC = 70^\circ$ and $\angle BCD = 70^\circ$ , diagonal $\overline{AC}$ bisects $\angle DAB$ , and diagonal $\overline{BD}$ bisects $\angle CDA$ .

Which choice is the measure of $\angle AEB$ ?

For a trapezoid with parallel bases, first convert the given lower interior angles into upper interior angles using supplementary consecutive interior angles. Apply any angle-bisector information next, then focus on a triangle containing the requested intersection angle. Finding its other two interior angles is often more reliable than trying to judge which angle at the diagonal intersection is intended.

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Use the parallel bases

Use the parallel lines to relate each given lower interior angle to the upper interior angle on the same leg of the trapezoid.

Split the upper angles

Each diagonal bisects one upper interior angle. Determine the angle that each relevant diagonal makes with a parallel base.

Use triangle $AEB$

Find the angles at $A$ and $B$ in triangle $AEB$ , then use the fact that a triangle’s interior angles sum to $180^\circ$ .

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Find an upper angle

Enter t=180-70. This represents either upper interior angle of the trapezoid.

Find the two angles in triangle $AEB$

Enter h=t/2 for the angle at $A$ , then enter b=70-h for the angle at $B$ .

Calculate the remaining triangle angle

Enter 180-h-b. The displayed value is the measure of $\angle AEB$ .

Step-by-step Explanation

Find the upper interior angles

Since $a \parallel b$ , each leg of the trapezoid is a transversal. Therefore, the consecutive interior angles along each leg are supplementary:

\angle DAB = 180^\circ - 70^\circ = 110^\circ

and

\angle CDA = 180^\circ - 70^\circ = 110^\circ.

Use the diagonal bisectors

Because $\overline{AC}$ bisects $\angle DAB$ ,

\angle EAB = \angle CAB = \frac{110^\circ}{2} = 55^\circ.

Because $\overline{BD}$ bisects $\angle CDA$ ,

\angle ADB = \frac{110^\circ}{2} = 55^\circ.

Since $\overline{AD} \parallel \overline{BC}$ , $\angle DBC = 55^\circ$ as well.

Find the angle in triangle $AEB$

At $B$ , the full angle $\angle ABC$ is $70^\circ$ . Since $\angle DBC=55^\circ$ and $E$ lies on $\overline{BD}$ ,

\angle ABE = 70^\circ - 55^\circ = 15^\circ.

Now use the angle sum of triangle $AEB$ :

\begin{aligned} \angle AEB &= 180^\circ - 55^\circ - 15^\circ \\ &= 110^\circ. \end{aligned}

Therefore, the measure of $\angle AEB$ is $110^\circ$ .

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Step-by-step Explanation

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Step-by-step Explanation