Question

In a $ \vartriangle PQR $ , $ P{R^2} - P{Q^2} = Q{R^2} $ and $ M $ is a point on side $ PR $ such that $ QM \bot PR $ , prove that $ Q{M^2} = PM \times MR $ .

Answer 1

Hint: Given that in a $ \vartriangle PQR $ ,
 $ P{R^2} - P{Q^2} = Q{R^2} $ and $ QM \bot PR $
We have to prove that $ Q{M^2} = PM \times MR $
Since, $ P{R^2} - P{Q^2} = Q{R^2} $
 $ \Rightarrow P{R^2} = P{Q^2} + Q{R^2} $
From Pythagoras theorem, we know that for a right angled triangle sum of squares of opposite side and adjacent side is equal to the square of hypotenuse side.
So from the above definition we can say that the $ \vartriangle PQR $ is a right angled triangle.
Also, it is given that $ M $ is a point on side $ PR $ such that $ QM $ is perpendicular to $ PR $

Using the concept of similar triangles we can solve this problem.

Complete step-by-step answer:

Given $ P{R^2} - P{Q^2} = Q{R^2} $
 $ \Rightarrow P{R^2} = P{Q^2} + Q{R^2} $
Therefore the $ \vartriangle PQR $ is a right angled triangle at $ Q $ .
Given $ QM \bot PR $
From figure,
In $ \vartriangle QMR $ ,
 $ \angle MQR + \angle M + \angle R = {180^ \circ } $
We know that $ \angle M = {90^ \circ } $
 $ \therefore \angle MQR = {180^ \circ } - {90^ \circ } - \angle R $
 $ \Rightarrow \angle MQR = {90^ \circ } - \angle R $
In $ \vartriangle PQR $ ,
 $ \angle P + \angle Q + \angle R = {180^ \circ } $
 $ \angle Q = {90^ \circ } $
 $ \therefore \angle P = {90^ \circ } - \angle R $
Now in $ \vartriangle QMR $ and $ \vartriangle PMQ $
 $ \angle M = \angle M $ (Both being $ {90^ \circ } $ )
 $ \angle MQR = \angle QPM $ (Both are equal to $ \left( {{{90}^ \circ } - \angle R} \right) $ )
From AA similarity criteria theorem, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are said to be similar.
 $ \therefore \vartriangle QMR \sim \vartriangle PMQ $
Now according to the theorem of area of similar triangles, the ratio of areas of two similar triangles is equal to the square of ratio of their corresponding sides
 $ \therefore \dfrac{{ar\left( {\vartriangle QMR} \right)}}{{ar\left( {\vartriangle PMQ} \right)}} = \dfrac{{{{\left( {QM} \right)}^2}}}{{{{\left( {PM} \right)}^2}}} $
 $ \Rightarrow \dfrac{{\dfrac{1}{2} \times RM \times QM}}{{\dfrac{1}{2} \times PM \times QM}} = \dfrac{{{{\left( {QM} \right)}^2}}}{{{{\left( {PM} \right)}^2}}} $ ( $ \because $ area of triangle = $ \dfrac{1}{2} $ $ \times $ base $ \times $ height)
On cancellation we get,
 $ Q{M^2} = PM \times RM $
Hence proved.

Note: AA similarity criteria theorem states that “if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are said to be similar”.
Since it is known that $ \vartriangle QMR \sim \vartriangle PMQ $ the ratio of their corresponding sides are also equal
 $ \dfrac{{QM}}{{PM}} = \dfrac{{RM}}{{QM}} = \dfrac{{QR}}{{PQ}} $
Therefore, in $ \vartriangle QMR $ and $ \vartriangle PMQ $
 $ QM $ and $ PM $ are corresponding sides.
Also the Area of similar triangles theorem states that “The ratio of areas of two similar triangles is equal to the square of ratio of their corresponding sides”.