In the figure, we join PA and PD. Draw perpendiculars on AB and CD from P as PM and PN respectively.
∠AEP = ∠ DEP (given)
So, ∠PEN = ∠PEM (M and N are points on line AE and ED respectively)
In ΔPEN and ΔPEM,
∠PNE = ∠PME = 90°
∠PEN = ∠PEM
PE = PE (common)
Therefore, Δ PEN ≅ ΔPEM (by AAS congruence)
∴ PN = PM (by CPCT)
We know that The chords of a circle equidistant from the center of a circle are congruent.
So, AB = CD.
Rate this question :