Question

In a quadrilateral ABCD, side BC || side AD, side AB = side DC. If $\angle$A = 72${}^{\circ }$ then find the measures of $\angle$B and $\angle$D.

Answer 1

Construction: Draw seg CQ $\perp$ side AD.
Draw seg BP $\perp$ side AD.

$\angle$A = 72${}^{\circ }$ [Given]

In ABCD, side AD = side BC [Given]
AB is their transversal.
$\angle$A + $\angle$B = 180${}^{\circ }$ [Interior angles on same side of transversal]
${72}^{\circ }+\angle B={180}^{\circ }$
$\angle B={180}^{\circ }–{72}^{\circ }$
$B={108}^{\circ }$

In $△$APB and $△$DQC,
AB $\cong$ DC [Given]
$\angle$APB = $\angle$DQC [Each angle is of measure 90${}^{\circ }$]
BP $\cong$ CQ [Perpendicular distance between two parallel lines]
By hypotenuse side test
$△$APB $\cong$ $△$DQC
$\angle$A = $\angle$D [c. p. c. t.]
$\angle$D = 72${}^{\circ }$

Hence measure of B and D are 108${}^{\circ }$ and 72${}^{\circ }$ respectively.