Question

Question 8
ABCD is a quadrilateral in which AB || DC and AD = BC. Prove that $\angle A=\angle Band\angle C=\angle D$.

Answer 1

Given ABCD is a quadrilateral such that AB || DC and AD = BC

Construction Extend AB to E and draw a line CE parallel to AD.

Proof Since, AD||CE and transversal AE cuts them at A and E, respectively.

$\therefore$ $\angle A+\angle E={180}^{\circ }$ [since, sum of cointerior angles is ${180}^{\circ }$]

$\Rightarrow$ $\angle A={180}^{\circ }-\angle E$

Since , AB || CD and AD || CE

So, quadrilateral AECD is a parallelogram.

$\Rightarrow AD=CE\Rightarrow BC=CE[\because AD=BC,given]$

Now, in $\Delta BCECE=BC\left[provedabove\right]$

$\Rightarrow \angle CBE=\angle CEB$ [opposite angles of equal side are equal]

$\Rightarrow {180}^{\circ }-\angle B=\angle E[\because \angle B+\angle CBE={180}^{\circ }]$

$\Rightarrow {180}^{\circ }-\angle E=\angle B$ . . . . . . . . (ii)
From Eqs. (i) and (ii) $\angle A=\angle B$ Hence proved.