3360.11 – Mystery Trapezoid

In the figure, $AB$ is parallel to $CD$ , and the measure of $\angle D$ is twice that of angle $B$ . What is $AB$ in terms of $a$ and $b$ ?

$\frac{a}{b} + 2b$
$\frac{3b}{2} + \frac{3a}{4}$
$2a - b$
$4b - \frac{a}{2}$
$a + b$

Solution

Draw $DE$ to bisect $\angle D$ . Then $\angle D1 = \angle D2 = \angle B$ . Now $\angle B + \angle C - 180$ , so $\angle D2 + \angle C = 180$ and $ED \parallel BC$ . So $BCDE$ is a parallelogram and $EB = b$ . Next,

\begin{aligned} \angle E1 = \angle B &\leadsto& \angle E1 = \angle D1 \\ &\leadsto& \triangle AED \text{ is isosceles} \\ &\leadsto& AE = a. \end{aligned}

Finally, $AB = AE + EB = a + b$ . The answer is e.