If a pair of opposite sides of a cyclic quadr | vedclass

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(N/A) $ABCD$ is a cyclic quadrilateral in which one pair of opposite sides $AB = DC$. We have to prove that diagonal $AC = $ diagonal $BD$.In $\Delta

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(N/A) $ABCD$ is a cyclic quadrilateral in which one pair of opposite sides $AB = DC$. We have to prove that diagonal $AC = $ diagonal $BD$.
In $\Delta AOB$ and $\Delta DOC$,we have:
$\angle 1 = \angle 3$ [Angles in the same segment of the circle are equal]
$AB = DC$ [Given]
$\angle 2 = \angle 4$ [Angles in the same segment of the circle are equal]
Therefore,$\Delta AOB \cong \Delta DOC$ [By $ASA$ congruence rule]
Therefore,$AO = OD$ [$CPCT$] $...(1)$
And $OC = BO$ [$CPCT$] $...(2)$
Now,adding $(1)$ and $(2)$,we get:
$AO + OC = BO + OD$
$\Rightarrow AC = BD$
Hence,proved.

If a pair of opposite sides of a cyclic quadrilateral are equal,prove that its diagonals are also equal.

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