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

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Solution

It is given that is isosceles and _.

We have to prove that .

It is given that is an isosceles triangle, so AC = BC.

Now,

(Given)

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

$∠ C A B = ∠ C B A (Equal sides have equal angles opposite to them) \Rightarrow 180 ° - ∠ C A P = 180 ° - ∠ C B Q \Rightarrow ∠ C A P = ∠ C B Q$

Hence, (SAS Similarity)

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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 $Δ A P C \sim Δ B C Q$ .

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 $Δ A C P \sim Δ B C Q$ .

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

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

△

ABC, the base AB is produced both the ways to P and Q such that

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

. Prove that

△ A P C \sim △ B C Q

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

It is given that is isosceles and . We have to prove that . It is given that is an isosceles triangle, so AC = BC. Now, (Given) ⇒APAC=BCBQ Also, ∠CAB=∠CBA Equal sides have equal angles opposite to them⇒180°-∠CAP=180°-∠CBQ⇒∠CAP=∠CBQ Hence, (SAS Similarity)

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