ABC is an isosceles triangle in which AB=AC which is circumscribed about a circle as shown in the figure. Show that BC is bisected at the point of contact.

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Detailed Solution

It is given that in an isosceles triangle AB=AC which is circumscribed about a circle.

As tangents drawn from an external point to a circle are equal in length.

So, we get

𝐴𝑃 = 𝐴𝑄 (tangents from A)

𝐵𝑃 = 𝐵𝑅 (tangents from B)

𝐶𝑄 = 𝐶𝑅 (tangents from C)

It is given that ABC in an isosceles triangle with sides

𝐴𝐵 = 𝐴𝐶

𝐴𝐵 − 𝐴𝑃 = 𝐴𝐶 − 𝐴𝑃

𝐴𝐵 − 𝐴𝑃 = 𝐴𝐶 − 𝐴𝑄

𝐵𝑅 = 𝐶𝑄

Since 𝐵𝑅 = 𝐶𝑄 that implies BC is bisected at the point of contact.

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