Question

in the diagram below, $overline{ad}congoverline{ae}$, $overline{ba}paralleloverline{ce}$, $overline{cb}paralleloverline{da}$ and $mangle dae = 50^{circ}$. find $mangle bcd$. you may assume lines that appear straight are straight, but the figure is not otherwise drawn to scale.

Explanation:

Step1: Find angles in isosceles triangle ADE

Since $\overline{AD}\cong\overline{AE}$ and $\angle DAE = 50^{\circ}$, in $\triangle ADE$, using the angle - sum property of a triangle ($\angle ADE+\angle AED+\angle DAE=180^{\circ}$) and the fact that $\angle ADE=\angle AED$ (base - angles of an isosceles triangle), we have $\angle ADE=\angle AED=\frac{180^{\circ}-\angle DAE}{2}=\frac{180 - 50}{2}=65^{\circ}$.

Step2: Use properties of parallel lines

Because $\overline{CB}\parallel\overline{DA}$, then $\angle BCD+\angle ADC = 180^{\circ}$ (consecutive interior angles of parallel lines). And $\angle ADC = 180^{\circ}-\angle ADE=180 - 65=115^{\circ}$.

Step3: Calculate $\angle BCD$

From $\angle BCD+\angle ADC = 180^{\circ}$, we can find $\angle BCD=180^{\circ}-\angle ADC$. Substituting the value of $\angle ADC$, we get $\angle BCD = 115^{\circ}$.

Answer:

$115^{\circ}$