Hello prantik007 Originally Posted by
prantik007 From an external point P , two tangents PA & PB are drawn to a circle with center O. Prove that OP is perpendicular bisector of AB.
My thanks to sa-ri-ga-ma for his contribution so far. He is quite correct up to the point where he says:
Hence you can show that
D is mid point of AB
But I'm afraid that this might be a bit of a cop-out.
I think you also need to use the fact that $\displaystyle \angle DBP = 90^o-\angle OBD$. Then use the fact that $\displaystyle \angle BPD = \angle OBD$ to say that: $\displaystyle \angle DBP = 90^o-\angle BPD$
$\displaystyle \Rightarrow \angle BDP = 90^o$ (angle sum of $\displaystyle \triangle BDP$)
Now we have established that $\displaystyle AB \perp OP$, it's very easy to show that $\displaystyle \triangle$'s $\displaystyle AOD$ and $\displaystyle BOD$ are congruent, and hence that $\displaystyle AD = DB$.
Can you fill in the details now?
Grandad