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Distance and Displacement

PQR is a tria...

Question

PQR is a triangle right angled at P and M is a point on QR such that PM $⊥$ QR. Show that $P M^{2} = Q M . M R$ .

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Solution

In $△ P M R$ ,
By Pythagoras theorem,
$(P R)^{2} = (P M)^{2} + (R M)^{2} . . . . . . . (1)$
In $△ P M Q$ ,
By Pythagoras theorem,
$(P Q)^{2} = (P M)^{2} + (M Q)^{2} . . . . . . . (2)$
In $△ P Q R$ ,
By Pythagoras theorem,
$(R Q)^{2} = (R P)^{2} + (P Q)^{2} . . . . . . . . (3)$
$∴$ $(R M + M Q)^{2} = (R P)^{2} + (P Q)^{2}$
$∴$ $(R M)^{2} + (M Q)^{2} + 2 R M . M Q = (R P)^{2} + (P Q)^{2} . . . . (4)$
Adding 1) and 2) we get,
$(P R)^{2} + (P Q)^{2} = 2 (P M)^{2} + (R M)^{2} + (M Q)^{2} . . . (5)$
From 4) and 5) we get,
$2 R M . M Q = 2 (P M)^{2}$
$∴$ $(P M)^{2} = R M . M Q$

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