Question 13·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

Because $AD=DC$, triangle $ADC$ is isosceles: it has two equal sides. Equal sides face equal angles, so the angle across from $AD$ equals the angle across from $DC$. Call their shared measure $a$:

Since $D$ lies on $BC$, the ray from $C$ through $D$ points toward $B$. So these are the same angle at $C$:

Now $AB=AC$. In isosceles triangle $ABC$, the base angles across from those equal sides match. So the angle at $B$ also measures $a$:

Because $D$ is between $A$ and $E$, $\angle CDE$ lies outside triangle $ADC$. The exterior-angle theorem says it equals the two inside angles that do not touch $D$. Those are the equal angles at $A$ and $C$. Add them:

The given exterior angle is $6x+26$, and $a=4x+4$. Put those angle measures into $m\angle CDE=2a$:

Type $6x_{1}+26\sim2(4x_{1}+4)$. Desmos uses $x$ for graphing, so use $x_{1}$ for the unknown and $\sim$ for a regression that finds it. Under PARAMETERS, Desmos shows $x_{1}=9$. That is $x$, not an angle measure.

Type $4x_{1}+4$ on the next Desmos line. It shows $40$, so $a=40^\circ$. This is the measure of $\angle CAD$ and the two base angles, not yet $\angle BAD$.

A triangle angle sum is $180^\circ$. At $A$, $\angle BAD$ and $\angle CAD$ make the whole angle; the angles at $B$ and $C$ each measure $a$. Count those three copies of $a$:

Subtract $3a$ to isolate the requested angle: $m\angle BAD=180-3a$. Type $180-3(40)$ on the next Desmos line. It shows $60$, so $\angle BAD$ measures $60^\circ$. Grid in 60.