Question
If two non - parallel sides of a trapezium are equal, prove that it is cyclic.
✓
Answer
Given: Non parallel sides $AD$ and $BC$ of a trapeziumare equal.
To prove : $ABCD$ is a cyclic trapezium.
Construction: Draw $DK \perp AB$ and $CP \perp AB$
Proof: In it $\Delta ADK$ and $\Delta BCP$
$AD=BC$
$DK=CP$ [Distance between || sides]
$DKA=CBP$ [ each $90{}^\circ$ ]
By RHS criteria of congruency,
$\triangle DKA \cong \triangle CPB$
$\angle A=\angle B $
$\angle ADK=\angle BCP\\ \angle ADK+90{}^\circ =\angle BCP+90{}^\circ $
$\angle ADC=\angle BCD$
$\angle D=\angle C ................(ii)$
$\angle A+\angle B+\angle C+\angle D=360{}^\circ [\angle A= \angle B , \angle C= \angle D]$
$\angle B+\angle B+\angle D+\angle D=360{}^\circ \\ \angle B+\angle D=180{}^\circ $
Hence, $ABCD$ is a cyclic trapezium.
To prove : $ABCD$ is a cyclic trapezium.
Construction: Draw $DK \perp AB$ and $CP \perp AB$
Proof: In it $\Delta ADK$ and $\Delta BCP$
$AD=BC$
$DK=CP$ [Distance between || sides]
$DKA=CBP$ [ each $90{}^\circ$ ]
By RHS criteria of congruency,
$\triangle DKA \cong \triangle CPB$
$\angle A=\angle B $
$\angle ADK=\angle BCP\\ \angle ADK+90{}^\circ =\angle BCP+90{}^\circ $
$\angle ADC=\angle BCD$
$\angle D=\angle C ................(ii)$
$\angle A+\angle B+\angle C+\angle D=360{}^\circ [\angle A= \angle B , \angle C= \angle D]$
$\angle B+\angle B+\angle D+\angle D=360{}^\circ \\ \angle B+\angle D=180{}^\circ $
Hence, $ABCD$ is a cyclic trapezium.
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