Question

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.

Answer 1

SOLUTION :  

Given : ΔABC is isosceles ∆ and AP x BQ = AC²

To prove : ΔAPC∼ΔBCQ.

Proof :  

ΔABC is an isosceles triangle AC = BC.

AP x BQ = AC²  (given)

AP x BQ = AC x AC

AP x BQ = AC x BC

AP/BC = ABQ……….(1).

Since, AC = BC

Then, ∠CAB = ∠CBA    

(angles opposite to equal sides are EQUAL)

180° – ∠CAP = 180° – ∠CBQ

∠CAP = ∠CBQ ………..(2)

In ∆APC &  ΔBCQ

AP/BC = AC/BQ [From equation 1]

∠CAP = ∠CBQ [From equation 2]

ΔAPC∼ΔBCQ  (By SAS similarity criterion)

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Hey,

In isosceles triangle ABC:-

AC = BC 

AP * BQ = AC^2 

prove that: ΔACP ~ ΔBCQ => prove similarity
 
∠CAB = ∠CBA => base angles of isosceles triangle 

∠CAB + ∠CAP = 180° => linear pair
 
∠CAP = 180° - ∠CAB => eq-1 

also: 

∠CBA + ∠CBQ = 180° => linear pair 

∠CBQ = 180° - ∠CBA => eq-2 

from 1 and 2 we get: 

∠CAP = ∠CBQ 

given: 

AP * BQ = AC^2, then: 

AP/AC = AC/BQ 

also: 

AP/AC = BC/BQ, thus: 

AP = BQ 

therefore by SAS: 

ΔACP ~ ΔBCQ.

HOPE IT HELPS YOU:-))