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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