Question

question is in pic, fast guys

Answer 1

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:-))

Answer 2

$\huge{hey}$

$\huge{ \mathfrak{here \: is \: answer}}$


$\bf{given}$

AB=AC=>/_B=/_C

$\sf{180 \degree - \angle{b} = 180 \degree - \angle{c}}$


$\sf{so \: \angle{abp} = \angle{acq}}$

also given that

BP*CQ=AB*AB
$\sf{ \frac{bp}{ba} = \frac{ab}{qc}}$

$\sf{ \frac{bp}{ca} = \frac{ab}{qc}}$
(since, ab=ac)



$\sf{ \frac{bp}{ca} = \frac{ab}{qc} and \angle{pba} = \angle{qca}}$

hence


triangle BPA~ triangle CAQ


$\huge \boxed{ \ulcorner{hope \: it \: helps}}$


$\huge{ \mathfrak{ \mathbb{THANKS}}}$