Question

in the given figure ABC is an isosceles triangle with ac is equal to BC ab is produced on either side till P and Q respectively such that AP× BQ=AC^2.prove that angle PCA= angle CQB

Answer 1

Draw a rough figure and consider the following:-

In isosceles triangle ABC below: 
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 

Answer 2

Answer:

In isosceles triangle ABC below:  

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

Step-by-step explanation: