MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB

0 like 0 dislike
51 views
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 $AP=BP$
| 51 views

0 like 0 dislike
<OAP=OBP= 90 DEGREES tangents

line 0A is a bisector of <APB

Therefore AP=BP ,equiangular
by Diamond (43,618 points)

0 like 0 dislike