SAT Lines, Angles & Triangles Question 36 | Free SAT Prep

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Question 36·Hard·Lines, Angles, and Triangles

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In the figure, $\overline{AC}$ and $\overline{BD}$ intersect at point $P$ . The segments satisfy $AP = BP$ and $CP = DP$ . The measure of $\angle APB$ is $40^\circ$ , and the measure of $\angle PBC$ is $30^\circ$ . Which choice is the measure, in degrees, of $\angle DAB$ ?

When a quadrilateral’s diagonals intersect and you are given equal segments from the intersection point, try forming triangles that use one segment from each diagonal and a vertical angle at the intersection. Proving those triangles congruent often lets you “move” a known angle to a new location. Then check for an isosceles triangle (from equal sides like $AP=BP$ ) to compute missing base angles, and finally add or subtract angles that meet at the vertex you are asked about.

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Look for vertical angles at the intersection

Angles formed by $\overline{AC}$ and $\overline{BD}$ at $P$ come in vertical pairs. Identify a pair that belongs to triangles you can compare.

Use the tick marks as side equalities

You are given two pairs of equal segments: $AP = BP$ and $CP = DP$ . Try to use them with a vertical angle to prove two triangles congruent (SAS).

Split $\angle DAB$ using the diagonal

Point $P$ lies on $\overline{AC}$ . Think of $\angle DAB$ as the sum of the angle from $AD$ to $AC$ and the angle from $AC$ to $AB$ .

Desmos Guide

Store the given angles

Enter a=40 and b=30 to represent $\angle APB$ and $\angle PBC$ .

Compute the base angle in the isosceles triangle

Enter (180-a)/2 to compute the base angle of isosceles triangle $APB$ (this is $\angle PAB$ ).

Add the two pieces at vertex $A$

Enter b+(180-a)/2. The value displayed is the measure of $\angle DAB$ .

Step-by-step Explanation

Prove a triangle congruence at $P$

Consider triangles $\triangle APD$ and $\triangle BPC$ .

$AP = BP$ (given)
$DP = CP$ (given)
$\angle APD = \angle BPC$ because they are vertical angles formed by lines $\overline{AC}$ and $\overline{BD}$ .

So $\triangle APD \cong \triangle BPC$ by SAS.

Transfer the given $30^\circ$ angle to vertex $A$

From $\triangle APD \cong \triangle BPC$ , corresponding angles are equal. The angle at $B$ in $\triangle BPC$ corresponds to the angle at $A$ in $\triangle APD$ , so

\angle PBC = \angle PAD.

Given $\angle PBC = 30^\circ$ , it follows that $\angle PAD = 30^\circ$ .

Use the isosceles triangle $APB$

In $\triangle APB$ , $AP = BP$ , so the base angles at $A$ and $B$ are equal.

The vertex angle is $\angle APB = 40^\circ$ , so the two base angles sum to $180^\circ - 40^\circ = 140^\circ$ . Each base angle is

\angle PAB = \angle PBA = \frac{140^\circ}{2} = 70^\circ.

Combine the two angles at $A$

At vertex $A$ , ray $\overline{AP}$ lies on diagonal $\overline{AC}$ , splitting $\angle DAB$ into $\angle DAP$ and $\angle PAB$ .

So,

\angle DAB = \angle DAP + \angle PAB = 30^\circ + 70^\circ = 100^\circ.

Therefore, the correct answer is $100$ .

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Step-by-step Explanation

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Strategy

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Step-by-step Explanation