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 $Δ A C P \sim Δ B C Q$ .

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Solution

c

Given, an isosceles triangle ABC having base AB and AC = BC.

Also, $A P \times B Q = A C^{2} \Rightarrow \frac{A P}{A C} = \frac{A C}{B Q} \Rightarrow \frac{A P}{A C} = \frac{B C}{B Q}$

To prove that $Δ A C P \sim Δ B C Q$ .

Proof :

Given, AC = BC ........ (2)

⇒ ∠CAB = ∠CBA [Angles opposite to equal sides are equal]

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

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

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

⇒ ∠CAP = ∠CBQ

Now, In ΔAPC and ΔCBQ, we have

∠CAP = ∠CBQ

$\frac{A P}{A C} = \frac{B C}{B Q}$

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

Hence Proved

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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.

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 $Δ A C P \sim Δ B C Q$ .