Question 152·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

Because $DE\parallel AB$, line $AD$ crosses both parallel lines. The alternate interior angles $\angle ADE$ and $\angle BAD$ are equal, so $m\angle BAD=m\angle ADE=(3x+8)^\circ$. The parallel marking, not how the lines look, lets you carry the measure to $A$.

$AD$ bisects $\angle BAC$, meaning it divides that angle into two equal parts. Since $\angle BAD$ measures $(3x+8)^\circ$, the other half measures $m\angle CAD=(3x+8)^\circ$.

Since $AB=AC$, triangle $ABC$ is isosceles: it has two equal sides. Those sides and the equal angles at $A$ make triangles $ABD$ and $ACD$ mirror each other across $AD$. Their angles at $D$ are equal and together make a straight angle along $BC$. So the tip bisector is perpendicular to the base: $m\angle ADC=90^\circ$.

Because $D$ lies on $BC$, the marked angle at $C$ is also an exterior angle of triangle $ADC$: it lies outside that triangle where side $DC$ is extended. An exterior angle equals the two interior angles away from it, so: $7x+42=(3x+8)+90$. Use only one half of the angle at $A$, namely $\angle CAD$, in this triangle.

Type $7x_1+42\sim(3x_1+8)+90$. Replace $x$ with $x_1$ so Desmos solves for it rather than graphs it, and use $\sim$ in place of $=$. Under PARAMETERS, Desmos shows $x_1=14$. So the value of $x$ is $14$. Grid in 14.