Question
In the given figure, $ABCD$ is a quadrilateral in which $AD = BC$ and $\angle\text{ADC}=\angle\text{BCD}.$ Show that the points $A, B, C, D$ lie on a circle.
✓
Answer
$ABCD$ is a quadrilateral in which $AD = BC$ and $\angle\text{ADC}=\angle\text{BCD}.$
Draw $\text{DE}\perp\text{AB}$ and $\text{CF}\perp\text{AB}.$ In $\triangle\text{ADE}$ and $\triangle\text{BCF},$
we have: $\angle\text{ADE}=\angle\text{ADC}-90^\circ=\angle\text{BCD}-90^\circ=\angle\text{BCF}$
$[$given: $\angle\text{ADC}=\angle\text{BCD}]$ $AD = BC$ [given]
$\therefore\ \triangle\text{ADE}\cong\angle\text{BCF}$ [By $AAS$ congruency]
$\Rightarrow\ \angle\text{A}=\angle\text{B}$
Now, $\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ$
$\Rightarrow\ 2\angle\text{B}=2\angle\text{D}=360^\circ$
$\Rightarrow\ \angle\text{B}+\angle\text{D}=180^\circ$
Hence, $ABCD$ is a cyclic quadrilateral.
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