Plane Geometry

• May 7th 2010, 07:15 PM
fantasylo
Plane Geometry
Attachment 16755

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.
• May 7th 2010, 07:55 PM
sa-ri-ga-ma
$\angle{ACB}=90 degrees$
EC is perpendicular to OC. AD is perpendicular to AB.
Hence
$\angle{CBA} = \angle{ECA} = \angle{EAC}$
So EA = EC.
Similarly you can prove that EP = EC.
• May 7th 2010, 08:06 PM
Quote:

Originally Posted by fantasylo
Attachment 16755

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

$\angle ECA=\angle CBA$ -- 1
$\angle EAC=\angle CBA$(alternate segments)

so $\angle ECA=\angle EAC$

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

$\angle OAE=\angle OCE=90^o$ (tangents at A and C)

Hence we can conclude that OAEC is a square

$\angle BCA=90^o$ (angle in a semicircle)

$
\angle DCE+\angle ECA+\angle BCA=180^o
$

$\angle DCE+\angle ECA=90^o$ (angle of a line)

$\angle OAC+\angle BCA+\angle CBA =180^o$

From 1 , we see that $\angle CAB=\angle ECD$

$\angle AOC =\angle DEC =90^o$
so $\triangle ADC \cong \triangle CED$