Question 23·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

Call the requested angle $m\angle BDC=x^\circ$. Since $BC=CD$, triangle $BCD$ is isosceles: its equal sides face equal angles. Side $CD$ faces the angle at $B$, so $m\angle CBD=x^\circ$.

Since $AB=BD$, the angles opposite those sides in triangle $ABD$ are equal. They are the angles at $D$ and $A$. Because $C$ lies between $A$ and $D$, rays $DC$ and $DA$ point the same way, so $m\angle BDA=m\angle BDC=x^\circ$. That makes $m\angle BAD=x^\circ$ too.

Because $C$ is between $A$ and $D$, ray $BC$ splits $\angle ABD$ into $\angle ABC$ and $\angle CBD$. Add their measures:

The interior angles of triangle $ABD$ add to $180^\circ$. Its angles at $A$ and $D$ are both $x^\circ$, and its angle at $B$ is $(84+x)^\circ$. Add them:

Combine the $x$ terms:

Type $84+3x_1\sim180$ in Desmos. Here $x_1$ is the unknown $x$, and the regression sign $\sim$ tells Desmos to find the value that makes both sides equal. Under PARAMETERS, Desmos shows $x_1=32$. That's the angle at $D$, so $m\angle BDC=32^\circ$. Choice D.