Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
Text Solution
Verified by Experts
By pythagoras theorem,
`OA^2 = AP^2 + OP^2`
In `/_\BPO`
`OB^2 = BP^2 + OP^2`
from above we get
`AP = BP`
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