Question

If two intersecting chords of a circle make equal angles with diameter passing through their point of intersection, prove that the chords are equal.

Answer 1

Let ‘O’ be the centre of the circle. 

PQ is a diameter.

\(\overline {AB}\)  and \(\overline {CD}\) are two chords meeting at E, a point on the diameter. 

∠AEO = ∠DEO 

Drop two perpendiculars OL and OM from ‘O’ to 

AB and CD; 

Now in ΔLEO and ΔMEO 

∠LEO = ∠MEO [given] 

EO = EO [Common] 

∠ELO = ∠EMO [construction 90°] 

∴ ΔLEO ≅ ΔMEO 

[ ∵ A.A.S. congruence] 

∴ OL = OM [CPCT]

i.e., The two chords  \(\overline {AB}\)  and \(\overline {CD}\) are at equidistant from the centre ‘O’. 

∴ AB = CD 

[∵ Chords which are equidistant from the centre are equal] 

Hence proved.