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In P Q R ,\ \ Q M|P R and P R^2-P Q^2=Q...

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

Text Solution

Verified by Experts

Given that
`PR^(2) - PQ^(2) = QR^(2) and QM bot PR `
`PR^(2) = PQ^(2) + QR^(2)`
`anglePQR= 90^(@)` ( converse of Pythagoras theorem)
In `triangleQMR and trianglePMQ`
`angleQMR = anglePMQ " " (each 90^(@)))`
` angleMQR= angleMPQ " " (each 90^(@)=angleR)`
` triangleQMR ~ anglePMQ` (AA similarity)
` Rightarrow " " (area (triangleQMR))/(area(trianglePMQ))= (QM^(2))/(PM^(2))`
`(1/2xxQMxxRM)/(1/2xxQMxxPM)= (QM^(2))/(PM^(2))`
` RM xx PM = QM^(2)`
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