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Consider a circle, with centre $O$. Choose a point $P$ outside the circle. Draw two tangents to the circle from point $P$, that meet the circle at $A$ and $B$. Draw lines $OA$, $OB$ and $OP$.

Prove that $AP=BP$
in Mathematics by Wooden (2,002 points) | 44 views

1 Answer

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<OAP=OBP= 90 DEGREES tangents

line 0A is a bisector of <APB

Therefore AP=BP ,equiangular
by Diamond (42,256 points)

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