Question

ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect at P. If AB = DC, prove that:

1. ΔΡΑΒ ≅ ΔPDC 
2. PA = PD and PC = PB

Answer 1

Given:

ABCD is a cyclic quadrilateral

Diagonals AC and BD intersect at P

AB = DC

(i) Prove ΔPAB ≅ ΔPDC

Since ABCD is a cyclic quadrilateral, points A, B, C, D lie on a circle.

Consider angles ∠PAB and ∠PDC.

These are angles subtended by the same chord, AB and DC, respectively (because AB = DC).

By the property of angles in the same segment of a circle, ∠PAB = ∠PDC.

Similarly, angles ∠PBA and ∠PCD are equal for the same reason (angles in the same segment).

Also, side AB = DC (given). 

Triangles ΔPAB and ΔPDC have:

Side AB = DC (given)

∠PAB = ∠PDC 

∠PBA = ∠PCD 

Therefore, by the Angle-Side-Angle (ASA) criterion, ΔPAB ≅ ΔPDC.

(ii) Prove PA = PD and PC = PB

From the congruence of triangles ΔPAB and ΔPDC, corresponding parts are equal.

Hence, PA = PD and PB = PC.

By a property of cyclic quadrilaterals, if two chords, AB and CD, intersect at P inside the circle:

PA × PB = PC × PD

AB = DC, and from the congruence, PA = PD and PB = PC, which also satisfy this relationship.