In the figure, if AB = AC prove that BQ = QC.

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Solution

It is given that $A B = A C$ .

But from the given figure, we have $A B = A P + P B$ and $A C = A R + R C$

Therefore, $A P + P B = A R + R C$

But $A P = A R,$ $P B = B Q$ and $R C = Q C$ (tangents drawn from same external points are equal)

Therefore, $A P + B Q = A P + Q C$

Hence, $B Q = Q C$ .

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Δ

∠

⊥

⊥

B C^{2} = (A C \times C P + A B \times B Q)

△ A B C, ∠ A

P B ⊥ A C

Q C ⊥ A B .

A B \times A Q = A C \times A P

{B C}^{2} = (A C \times C P + A B \times B Q)

In the figure, if $A B = A C a n d ∠ B = ∠ C .$

Prove that BQ = CP.

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