Question

^Answer the question from the attachment^
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Answer 1

prove that
XZP ~ QZX
by AA
angle z =angle z
angle Q = angle x
LQ+LP =90
LP+LX = 90

Answer 2

Hey !!

Here is your answer..

↪Data :- ∆PQR is right angled triangle at Q.QX perpendicular PR,XY perpendicular RQ and XZ perpendicular PQ.

↪To Prove :- XZ ^2 = PZ × ZQ

↪proof :- In ∆ PZX and ∆ PXQ
angle P = angle P ......( common angles )
angle PZX = angle PXQ ..( right angles )

∆ PZX ~ ∆ PXQ .....( by AA criterion ) ( 1 )

Similarly, ∆ XZQ ~ ∆ PXQ ...... ( 2 )

From eq. ( 1 ) and eq. ( 2 )

∆ PZX ~ ∆ XZQ

$\frac{pz}{xz} = \frac{xz}{zq} \: \: \: \: .......(csst) \\ \\ {xz}^{2} = pz \times zq \: \: \: \: \: ....(proved)$

Hope it helps You...

THANKS
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