**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**