Question

From the endpoint of a diameter of a circle perpendicular drawn to a tangent of the same circle .show that their feet on the tangent are equidistant from the centre of the circle .​

Answer 1

Proved below.

Step-by-step explanation:

Given:

As shown in the figure below, circle with center O and AB is a diameter.

Line CD is tangent, touching the circle at E

AC ⊥CD and BD ⊥CD

As line CD is a tangent at point E and OE is radius,

OE⊥CD

∴ AC⊥CD and OE⊥CD and  BD⊥CD

⇒A║O║BD

By the property of intercepts made by 3 parallel lines,

$\frac{AO}{OB} =\frac{CE}{ED}$

But AO = OB            (radii of the same circle)

∴$\frac{AO}{OB} =\frac{CE}{ED}$ = 1

⇒ CE = ED

In ΔOEC and ΔOED

OE = OE     (common side)

CE = ED      (from above)

∠OEC = ∠OED = 90° (∵ OE ⊥ CD)

BY SAS congruency

ΔOEC ≅ ΔOED

OC = OD    (By CPCT)

Hence proved.