Q2 of 15 Page 86

Two chords intersect at a point on a circle and the diameter through this point bisects the angle between the chords. Prove that the chords have the same length.


EF and FG are two chord meet at point F on circle centred at D.


FH is the diameter through point F.


We have to prove that EF = FG.


We have joined EH and HG.


In ΔEHF and ΔGHF we have,


HF is the common side.


∠EFH = ∠ GFH [HF bisects the angle between two chords]


∠HEF = ∠HGF = 90° [both are angle inscribed on semicircle]


∴∆EHF≅∆GHF [RHS congruency]


∴ EF = FG [similar side of congruent triangle]


∴ The chords have same length.


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