Exercise 181

Quadrilateral $ABCD$ is inscribed in a circle with $\angle BAC=69^\circ,\angle ADB=42^\circ, AD=4, BC=8$. What is $AC$?

Solution

$\angle ACB=\angle ADB=42^\circ$ (same arc $AB$), so $\angle ABC=180^\circ-69^\circ-42^\circ=69^\circ=\angle BAC$.
Since $\angle BAC=\angle ABC$, triangle $ABC$ is isosceles with $AC=BC$ (sides opposite equal angles).
$AC=BC=\boxed8$.