Question 344·Hard·Math

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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 $\angle DAB$ ?

Crossing diagonals with paired equal sides point to triangle congruence. The opposite angles at the crossing are equal, giving you the included angle for side-angle-side (SAS). Carry a given angle to the corner you need, then use the equal sides of another triangle to find the remaining part of that corner. Don’t stop after finding only one part.

Hints

Hint 1
Opposite angles where two lines cross are vertical angles, and they are equal. Which angles at $P$ sit between the two pairs of marked equal sides?
Hint 2
SAS congruence uses two pairs of equal sides and an included angle, the angle between those sides. Pair $AP$ with $BP$ and $DP$ with $CP$ . Which angle at $A$ then matches the given angle at $B$ ?
Hint 3
In an isosceles triangle, equal sides face equal angles. Use $AP=BP$ to find the angle at $A$ in triangle $APB$ . Is that the whole angle the question asks for?

Step-by-step

Match the triangles, then add the angles
Step 1
Find the equal angles at the crossing
The diagonals cross at $P$ , so $\angle APD$ and $\angle BPC$ sit opposite each other. Such angles are called vertical angles, and they’re equal: $m\angle APD=m\angle BPC$ .
Step 2
Use the marked sides for SAS
The equal angles at $P$ are included angles: each lies between two marked sides. Since $AP=BP$ and $DP=CP$ , side-angle-side (SAS) gives $\triangle APD\cong\triangle BPC$ . Congruent triangles have the same size and shape. Check that the equal angle lies between the paired sides before using SAS.
Step 3
Carry the given angle to A
In the congruent triangles, $A$ matches $B$ : side $AP$ matches $BP$ , while $DP$ matches $CP$ . So the matching angles are equal: $m\angle DAP=m\angle PBC=30^\circ$ . The angle at $C$ does not match the angle at $A$ .
Step 4
Identify the other angle at A
Since $AP=BP$ , triangle $APB$ is isosceles, meaning it has two equal sides. Equal sides face equal angles, so $m\angle PAB=m\angle ABP$ .
Step 5
Find that equal angle
The angles of a triangle add to $180^\circ$ . After the $40^\circ$ angle at $P$ , the two equal base angles share what remains. Type $\frac{180-40}{2}$ in Desmos; it displays $70$ . So $m\angle PAB=70^\circ$ .
Step 6
Add the two parts of the target angle
Ray $AP$ divides $\angle DAB$ into two neighboring angles. Add their measures:
$\begin{aligned}m\angle DAB&=m\angle DAP+m\angle PAB\\[1.4em]&=30^\circ+70^\circ\\[1.4em]&=100^\circ.\end{aligned}$
So $\angle DAB$ measures $100^\circ$ . Choice C.

Question 344·Hard·Math

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