Question

PQR is a right angled at P and M is a point on QR such that PM perpendicular to QR. Show that PM2=QM×MR.

Answer 1

here, we have to first prove that tr. pmr is similar to tr. qmr. (both the smaller triangles inside PQR)

we know that pmr~pqr and pqm~pqr ( by a theorem, its there in the textbook)

=> pmr~pqm

now,

RM/PM = PM/QM (ratios of corresponding sides in similar triangles are equal)

by cross multipication, we get

RM x QM = PM x PM

PM2 = QM x MR

 

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