Question

In a triangle PQR,N is a point on PR such that QN is perpendicular to PR if PR.NR=QN^2 then prove that angle PQR= 90°​

Answer 1

Answer:

PN×NR=QN  

2

 

Given: PN=QN,QN=NR

In ΔPNQ and ΔQNR

∠PNQ=∠QNR=90  

o

 

PN=QN,QN=NR

ΔPNQ∼ΔQNR by S.A.S

∴∠PQN=∠QRN

In ΔQNR,

∠QNR+∠QRN+∠NQR=180  

o

 by angle sum property

⇒90  

o

+∠PQN+∠NQR=180  

o

 

⇒∠PQN+∠NQR=180  

o

−90  

o

=90  

o

 

⇒∠PQR=90  

o

 

Hence proved.

solution

Step-by-step explanation: