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