Question

$△PQR$ is right angled at $Q,QX\perp PR,XY\perp RQ$ and $XZ\perp PQ$ are drawn. Prove that $X{Z}^2=PZ\times ZQ$.

Answer 1

In ΔPZX and ΔPQR

∠P=∠P (∵common)

∠PZX=∠PQR=90° (given)

By AA similarity ΔPZX≈ΔPQR → (1)

In ΔXZQ and ΔPXQ

∠XZQ=∠PXQ (90°each)

∠ZQX=∠PQX (∵common)

By AA similarity, ΔXZQ≈ΔPXQ → (2)

In ΔPQR and ΔPXQ

∠PQR=∠PXQ (∵each 90°)

∠P=∠P (∵common)

By AA similarity, ΔPQR ≈ΔPXQ → (3)

From eq (1),(2) and (3) transitivity property

ΔPZX≈ΔXZQ

Hence, AA similarity postulates,

⇒ $\frac{PZ}{ZX}=\frac{ZX}{ZQ}\Rightarrow Z{X}^2=PZ\times ZQ$

Hence, proved