AB and AC are two equal chord of a circle, therefore the centre of the circle lies on the bisector of ∠BAC.
OA is the bisector of ∠BAC.
Again, the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle.
P divides BC in the ratio 6:6=1:1.
P is mid-point of BC.
OP ⊥ BC.
In △ ABP, by pythagoras theorem,
AB2=AP2+BP2
BP2=36−AP2 ....(1)
In △ OBP, we have
OB2=OP2+BP2
52=(5−AP)2+BP2
BP2=25−(5−AP)2 .....(2)
From 1 & 2, we get,
36−AP2=25−(5−AP)2
36=10AP
AP=3.6cm
Substitute in equation 1,
BP2=36−(3.6)2=23.04
BP=4.8cm
BC=2×4.8=9.6cm
