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In given figure , ABC is a triangle produced meets the circumcircle of `Delta ABC` at `Q`, prove that `CP=CQ` |
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Answer» In ` Delta ABP`, `AB=AP` (given) `angle1=angle2` (angles opposite to equal sides ) .......(1) But `angle2 =angle3 ` (vertically opposite angle) ......(2) `rArrangle1=angle3` [From (1) and (2)] .........(3) But `angle1=angle4` (angles of same segment) .........(4) `therefore angle3= angle4` [from (3)and (4)] `rArrCP=CQ` (side opposite to equal of `DeltaCPQ`) |
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