Question

In $\square$ABCD , side BC || side AD, side AB $\cong$ side DC If $\angle$A = ${72}^{°}$ then find the measure of $\angle$B, and $\angle$D.

Answer 1

side BC || side AD
Interior angles on the same side of the transversal are supplementary.
$\angle$A + $\angle$B = 180º
$\Rightarrow$${72}^{°}$ + $\angle$B = 180º
$\Rightarrow$$\angle$B = 180º − ${72}^{°}$
$\Rightarrow$$\angle$B = 108º
Construction: Draw BP || CD
So, BC || AD and BP || CD
$\Rightarrow$PBCD is a parallelogram
$\Rightarrow$CD $\cong$ BP
Now CD = BP and CD = AB
$\Rightarrow$BP = AB
$\Rightarrow$$\angle \mathrm{BAP}=\angle \mathrm{BPA}=72°$
BP || CD so,
$\angle \mathrm{CDP}=\angle \mathrm{BPA}=72°\left(\mathrm{corresponding}\mathrm{angles}\mathrm{are}\mathrm{equal}\right)$
$\angle$B = 108º
$\angle$D = 72º