In the given figure,
is a quadrilateral in which
and
Show that the points
lie on a circle.

In ΔADE and ΔBCF,
AD = BC
∠AED = ∠BFC
∠ADE = ∠BCF [∠ADC - 90° = ∠BCD - 90°]
∴ ΔADE ≅ ΔBCF
The Cross ponding parts of the congruent triangles are equal.
∴ ∠A = ∠B
Now,
∠A + ∠B + ∠C + ∠D = 360°
⇒ 2 ∠B + 2 ∠D = 360°
⇒ ∠B + ∠D = 180°
∴ ABCD is a cyclic quadrilateral.
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