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

MX² = AX.BX ...... Proved.

Step-by-step explanation:

See the attached diagram.

Since AB is the diameter of the circle, so ∠ AMB = 90°.

Now, from Pythagoras Theorem. we can write

MA² + MB² = AB² = (AX + XB)² = AX² + XB² + 2.AX.XB ............ (1)

Now, from right triangle Δ AMX, we can write

AX² = MA² - MX² ......... (2)

And from the right triangle, Δ BMX, we can write

XB² = MB² - MX² ........... (3)

Now, from equations (1), (2), and (3) we can write

MA² + MB² = (MA² - MX²) + (MB² - MX²) + 2.AX.XB

⇒ 2MX² = 2.AX.BX

⇒ MX² = AX.BX

Hence, proved.