# 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
\$\displaystyle \angle{ACB}=90 degrees\$
EC is perpendicular to OC. AD is perpendicular to AB.
Hence
\$\displaystyle \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

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

so \$\displaystyle \angle ECA=\angle EAC\$

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

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

Hence we can conclude that OAEC is a square

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

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

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

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

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