wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In the figure, an isosceles triangle ABC, with AB=AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.
495237_3bc3a3a44b5243228f25adea4e3cad3a.png

Open in App
Solution

Given: An isosceles ΔABC with AB=AC, circumscribing a circle.
To prove: P bisects BC.
Proof: AR and AQ are the tangents drawn from an external point A to the circle.
Therefore, AR=AQ (Tangents drawn from an external point to the circle are equal)
Similarly, BR=BP and CP=CQ.
It is given that in ΔABC, AB=AC.
AR+RB=AQ+QC
BR=QC(As AR=AQ)
BP=CP(As BR=BP and CP=CQ)
P bisects BC.
Hence, the result is proved.
565551_495237_ans_4a2eda4aaa394a6f8359320664514384.png

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Intersection between Secants
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon