In ` P Q R ,\ \ Q M_|_P R` and `PR^2-P Q^2=Q R^2`.Prove that `Q M^2=P MxxM R`

Explanation

Given in `DeltaPQR`, `PR^(2)-PQ^(2)=QR^(2) and QMbotPR`
To prove `QM^(2)=PMxxMR`
Proof Since, `PR^(2)-PQ^(2)=QR^(2)`
`rArr PR^(2)=PQ^(2)+QR^(2)`

So, `DeltaPQR` is right angled triangle at Q.
In `DeltaPQR` iss right angled triangle at Q.
In `DeltaQMR` and `DeltaPMQ`,
`angleM=angleM` [each `90^(@)`]
`angleMQR=angleQPM` [each equal to `90^(@)-'angleR`]
`DeltaQMR~DeltaPMQ` [by AAA similarity criterion]
Now, using property of area of similar triangles, we get
`(ar(DeltaQMR))/(ar(DeltaPMQ))=((QM)^(2))/((PM)^(2))`
`rArr (1/2xxRMxxQM)/(1/2xxPMxxQM)=((QM)^(2))/((PM)^(2))` [`because` areaa of triangle=`1/2xx`base`xx`height]
`rArr QM^(2)=PMxxRM` Hence proved.