In the figure, and intersect at point . The segments satisfy and . The measure of is , and the measure of is . Which choice is the measure, in degrees, of ?

Solution available with step-by-step explanation

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

00:00

Question 36·Hard·Lines, Angles, and Triangles

0 / 50

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.

Hints

Click me!

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 $100$ .

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation