Question

A semicircle is drawn on AB as diameter .Let X be a point on AB.From X a perpendicular XM is drawn on AB cutting the semicircle at M .prove that AX.XB=MX squre​

Answer 1

Answer:

Proved as below.

Step-by-step explanation:

i.) $MA^{2} + MB^{2} = AB^{2}$

ii.) $AX^{2} + MX^{2} = MA^{2}$

iii.) $XB^{2} + MX^{2} = MB^{2}$

from i)

$MA^{2} +MB^{2} = AB^{2} = AX^{2} + XB^{2} + 2AX.BX$

$AX^{2} +MX^{2} + XB^{2} + MX^{2} = AX^{2} + XB^{2} + 2AX.XB$

Therefore

$MX^{2} = AX.BX$