In the figure, an isosceles triangle $A B C,$ with $A B = A C,$ circumscribes a circle. Prove that the point of contact $P$ bisects the base $B C .$

Open in App

Solution

Given: An isosceles $Δ A B C$ with $A B = A C,$ circumscribing a circle.
To prove: $P$ bisects $B C .$
Proof: $A R$ and $A Q$ are the tangents drawn from an external point A to the circle.
Therefore, $A R = A Q$ (Tangents drawn from an external point to the circle are equal)
Similarly, $B R = B P$ and $C P = C Q$ .
It is given that in $Δ A B C$ , $A B = A C .$
$\Rightarrow A R + R B = A Q + Q C$
$\Rightarrow B R = Q C (A s A R = A Q)$
$\Rightarrow B P = C P (A s B R = B P a n d C P = C Q)$
$\Rightarrow$ $P$ bisects $B C .$
Hence, the result is proved.

Suggest Corrections

Similar questions

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

In Fig.4, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.

OR
In Fig.5, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.

A B = A C,

B E = E C .

A B C

A B = A C,

A B C

A B = A C

B C

Join BYJU'S Learning Program