Hello thereddevils 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:
$\displaystyle OC\cdot OD = 2\triangle OCD$ $\displaystyle = 2(\triangle OCE + \triangle ODE)$
$\displaystyle =2(\tfrac12pr + \tfrac12qr)$
$\displaystyle =(p+q)r$
$\displaystyle = \tfrac12CD\cdot AB$
Grandad