Question
ABCD is a quadrilateral in which AB || DC and AD = BC. Prove that $\angle\text{A}=\angle\text{B}$ and $\angle\text{C}=\angle\text{D}.$
✓
Answer
ABCD is a quadrilaateral such that AB || DC and AD = BC
Construction Extend AB to E and draw a line CE parallet to AD.
proos since, AD || CE and transverrsal AE cuts then at A and E repectively.
$\therefore\angle\text{A}+\angle\text{E}=180^\circ$ [since, sum of cointerios angles is 180°]
$\Rightarrow\ \angle\text{A}=180^\circ-\angle\text{E}\ ...(\text{i})$
So, quadrilateral AECD is a parallelogram.
$\Rightarrow\ \text{AD}=\text{CE}\Rightarrow\text{BC}=\text{CE}$ [$\because\ \text{AD}=\text{BC}$ given]
Now, in $\Delta\text{BCE}$ $\text{CE}=\text{BC}$ [proved above]
$\Rightarrow\ \angle\text{CBE}=\angle\text{CEB}$
[opposite angles of equal side are equal]
$\Rightarrow\ 180^\circ-\angle\text{B}=\angle\text{E}$ $[\because\angle\text{B}+\angle\text{CBE}=180^\circ]$
$\Rightarrow\ 180^\circ-\angle\text{E}=\angle\text{B}\ ...(\text{ii})$
From Eqs. (i) and (ii) $\angle\text{A}=\angle\text{B}$ Hence proved.
Construction Extend AB to E and draw a line CE parallet to AD.
proos since, AD || CE and transverrsal AE cuts then at A and E repectively.
$\therefore\angle\text{A}+\angle\text{E}=180^\circ$ [since, sum of cointerios angles is 180°]
$\Rightarrow\ \angle\text{A}=180^\circ-\angle\text{E}\ ...(\text{i})$
So, quadrilateral AECD is a parallelogram.
$\Rightarrow\ \text{AD}=\text{CE}\Rightarrow\text{BC}=\text{CE}$ [$\because\ \text{AD}=\text{BC}$ given]
Now, in $\Delta\text{BCE}$ $\text{CE}=\text{BC}$ [proved above]
$\Rightarrow\ \angle\text{CBE}=\angle\text{CEB}$
[opposite angles of equal side are equal]
$\Rightarrow\ 180^\circ-\angle\text{B}=\angle\text{E}$ $[\because\angle\text{B}+\angle\text{CBE}=180^\circ]$
$\Rightarrow\ 180^\circ-\angle\text{E}=\angle\text{B}\ ...(\text{ii})$
From Eqs. (i) and (ii) $\angle\text{A}=\angle\text{B}$ Hence proved.
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