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 $Δ A C P \sim Δ B C Q$ .
$[3]$

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Solution

Proof :

Given, AC = BC ........ (1)

⇒ ∠CAB = ∠CBA ........ (2)
[Angles opposite to equal sides are equal]

Again, ∠CAB + ∠CAP = 180° and ∠CBA + ∠CBQ = 180° [Both forming a linear pair]
$[1]$

⇒ ∠CAB = 180° – ∠CAP and ∠CBA = 180° – ∠CBQ

⇒ 180° – ∠CAP = 180° – ∠CBQ [using (2)]

⇒ ∠CAP = ∠CBQ
$[1]$

Now, In ΔAPC and ΔCBQ, we have

∠CAP = ∠CBQ

$\frac{A P}{A C} = \frac{A C}{B Q}$
from equation ...(1)
$\frac{A P}{A C} = \frac{B C}{B Q}$

$\Rightarrow Δ A C P \sim Δ B C Q$ .(SAS similarity)

Hence Proved
$[1]$

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Similar questions

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

Prove that $Δ A C P \sim Δ B C Q$ .

(3 Marks)

Q. 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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In an isosceles ΔABC, the base AB has produced both ways in P and Q such that AP×BQ=AC2. Prove that ΔACP∼ΔBCQ. [3]

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 $Δ A C P \sim Δ B C Q$ .
$[3]$