In a `Delta PQR`, N is a point on PR, such that QN`bot`PR. If PN`cdot`NR=`QN^(2)`, then prove that `anglePQR=90^(@)`.

Explanation

Given : `trianglePQR` in which `QN bot PR and PN xx NR = QN^(2)`
To prove , ` anglePQR = 90^(@)`
Proof : In `triangleQNP and triangleQNR`
`QN bot PR`
` angle2 = angle1 = 90^(@)`
` QN^(2)= NR xx NP`
`(QN)/(NR)= (NP)/(QN) or ( QN)/(NP) = (NR)/(QN)`
`triangleQNR ~ trianglePNQ` ( by SAS similarity criterion)
` angle3 = angleP` ..(1)
`angle2=angle1 = 90^(@)`
`angleR = angle4` ...(2)
In `trianglePQR` we have
`angleP +anglePQR+ angleR = 180^(@)` ( angle sum property of a trangle)
`angle3 + angle4+ angle3 + angle4 = 180^(@)` [ using (1) and (2) ]
`Rightarrow " " 2angle3+ 2angle4= 180^(@)`
`Rightarrow " " angle3+angle4 = 90^(@)`
`anglePQR = 90^(@)` Hence proved