# geometry (3)

• Nov 7th 2009, 06:42 AM
thereddevils
geometry (3)
In the diagram , AOB is a diameter of the circle with centre O . ACX , BDY and DEC are tangents to the circle at A , B and E respectively . Show that

(1) angle COD =90

(2) OC . OD = 1/2 AB . CD
• Nov 7th 2009, 07:22 AM
Hello thereddevils
Quote:

Originally Posted by thereddevils
In the diagram , AOB is a diameter of the circle with centre O . ACX , BDY and DEC are tangents to the circle at A , B and E respectively . Show that

(1) angle COD =90

(2) OC . OD = 1/2 AB . CD

Part (1) is like your previous question: Geometry (1). Here's the outline of the proof. I'll leave the details to you. Let $\displaystyle \angle ACO = x, \angle EDO = y$.

Then $\displaystyle \angle ECO = x$ and $\displaystyle \angle BDO = y$ (congruent $\displaystyle \triangle$'s)

Then look at the quadrilateral $\displaystyle ACDB$. Can you see why $\displaystyle 2x+2y = 180^o$? Hence $\displaystyle x+y=90^o$. Then look at $\displaystyle \triangle COD$, and you're there.

Part (2) Let $\displaystyle AC = CE = p,\,ED=BD = q,\, AO=EO=BO =r$.

Then look at the areas:

$\displaystyle OC\cdot OD = 2\triangle OCD$
$\displaystyle = 2(\triangle OCE + \triangle ODE)$

$\displaystyle =2(\tfrac12pr + \tfrac12qr)$

$\displaystyle =(p+q)r$

$\displaystyle = \tfrac12CD\cdot AB$