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

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In right angled triangles

ΔRPQ & ΔPMR,

∠RPQ = ∠PMR = 90°

∠PRQ = ∠PRM (Common angle)

Then ∠PQR = ∠MPR (sum of angles is 180°)

\(\therefore\) ΔRPQ \(\sim\) ΔPMR (By AAA similarity criteria)

\(\therefore\) \(\frac{MR}{PR}=\frac{PM}{PQ}=\frac{PR}{QR}\) .........(1)

Similarly, in right angled triangle

ΔPMQ & ΔQPR,

∠PMQ = ∠QPR = 90°

∠PQM = ∠PQR  (Common angle)

Then ∠QPM = ∠PRQ  (Sum of angle is 180°)

\(\therefore\) ΔPMQ \(\sim\) ΔQPR   (By AAA similarity criteria)

\(\therefore\) \(\frac{PQ}{QR}=\frac{PM}{PR}=\frac{QM}{PQ}\)  ..............(2)

From equation (1), we have

\(\frac{MR}{PR}=\frac{PM}{PQ}\)

\(\Rightarrow \frac{MR}{PM}=\frac{PR}{PQ}\) ..........(3)

From equation (2), we have

\(\frac{PM}{PR}=\frac{QM}{PQ}\)

\(\Rightarrow \frac{PM}{QM}=\frac{PR}{PQ}\)  ..........(4)

From equation (3) and (4), we have

\(\frac{PM}{QM}=\frac{QR}{PQ}=\frac{MR}{PM}\)

\(\Rightarrow \frac{PM}{QM}=\frac{MR}{PM}\)

\(\Rightarrow PM^2=QM.MR\)

Hence proved

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