In an isosceles $Δ A B C$ , the base $A B$ is produced both ways in $P$ and $Q$ , such that $A P \times B Q = A C^{2}$ . Prove that $Δ A C P \sim Δ B C Q$ .

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Solution

$A C = B C$

$A P * B Q = A C^{2}$

$∠ C A B = ∠ C B A$ (base angles of isosceles triangle)

$∠ C A B + ∠ C A P = 180 °$ (linear pair)
$∠ C A P = 180 ° - ∠ C A B$ ------(1)

also:

$∠ C B A + ∠ C B Q = 180 °$ (linear pair)

$∠ C B Q = 180 ° - ∠ C B A$ ------(2)

from 1 and 2 we get:
$∠ C A P = ∠ C B Q$
given: $A P * B Q = A C^{2}$

$⟹ \frac{A P}{A C} = \frac{A C}{B Q}$

$⟹ \frac{A P}{A C} = \frac{B C}{B Q}$

$⟹ A P = B Q$

Therefore by SAS: $Δ A C P \sim Δ B C Q$

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In an isosceles $Δ A B C$ , the base AB has produced both ways in P and Q such that $A P \times B Q = A C^{2}$ .

Prove that $Δ A C P \sim Δ B C Q$ .

In an isosceles $Δ A B C$ , the base AB is produced both the ways to P and Q such that $A P \times B Q = A C^{2}$ . Prove that $Δ A P C \sim Δ B C Q$ .

Q. In an isosceles ∆ ABC, the base AB is produced both ways to P and Q, such that AP × BQ = AC².
Prove that ∆ ACP ∼ ∆ BCQ.

In an isosceles

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△ A P C \sim △ B C Q

Q. In the adjoining figure P and Q are two points on equal sides AB and AC of an isosceles triangle ABC such that AP

=

AQ , prove that BQ

=

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In an isosceles ΔABC , the base AB is produced both ways in P and Q, such that AP×BQ=AC2. Prove that ΔACP∼ΔBCQ.

In an isosceles $Δ A B C$ , the base $A B$ is produced both ways in $P$ and $Q$ , such that $A P \times B Q = A C^{2}$ . Prove that $Δ A C P \sim Δ B C Q$ .