Question

Triangle PQR IS right angled at Q. QX IS PERPENDICULAR TO PR. XY IS PERPENDICULAR TO RQ. XZ IS PERPENDICULAR TO PQ are drawn prove that XZSqaure=PZ×ZQ

Answer 1

Answer:

The proof is explained below.

Step-by-step explanation:

Given Triangle PQR is right angled at Q. QX is perpendicular to PR. XY is perpendicular to RQ. XZ is perpendicular to PQ. we have to prove that $XZ^{2}=PZ\times ZQ$.

To prove above we have to prove that ΔPZX≈ΔXZQ

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, $\frac{PZ}{XZ}= \frac{ZX}{ZQ}$

⇒ $XZ^{2}=PZ\times ZQ$

Hence, proved

Answer 2

Step-by-step explanation:

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