Thread: Plane Geometry

The diagram shows a circle, centre O, with diameter AB. The point C lies on the circle. The tangent to the circle at A meets BC extended at D. The tangent to the circle at C meets the line AD at E.
How to prove that E is the mid-point of AD?

Thanks.

EC is perpendicular to OC. AD is perpendicular to AB.
Hence

So EA = EC.
Similarly you can prove that EP = EC.

Originally Posted by fantasylo

The diagram shows a circle, centre O, with diameter AB. The point C lies on the circle. The tangent to the circle at A meets BC extended at D. The tangent to the circle at C meets the line AD at E.
How to prove that E is the mid-point of AD?

Thanks.

hi

-- 1
(alternate segments)

so

and EA=EC (equal sides of an isosceles triangle)

(tangents at A and C)

Hence we can conclude that OAEC is a square

(angle in a semicircle)



(angle of a line)



From 1 , we see that

AD=EC (square)



so

ED=OC (corresponding angle)

Since OC=EA , ED=EA