Consider the figure as shown below:
Here, $ABC $ is a quadrant, a circle with centre $ O $ is inscribed in the quadrant such that it intersects the quadrant at $M,N $ and $P$;
Now,how can we prove that the straight line $BP$ will pass through the centre $O$ ?
My Idea:
If we can prove that $\angle POB =180°$ ,then we have proved that $BP $ will pass through the centre $O $
My Approach:
Join $OM$ and $ON $as follows:
Since,$MB$ is the tangent to the circle
$\implies \angle OMB =90 ° $
Similarly, $ \angle ONB =90 ° $
Also, as $ABC$ is a quadrant
$\implies \angle MBN =90 ° $
$\implies OMBN $ is a square
$\implies \angle NOB =45 ° $
So,we need to prove now that $\angle PON =135°$
So,we redraw the given figure as follows:
Here,$\angle MPN=45°$ as $ \angle MON =90°$
So, in $\triangle MPN$ ,if we can prove that $MP=NP$ ,then $\angle PMN= 67.5°$, which will give $\angle PON =135°$
$ \implies $ we need to prove $∆MPO \cong ∆NPO$
So, how to prove $\triangle MPO \cong \triangle NPO $ ?
Best Answer
You've made a big deal out of it.
Complete the circle from the quadrant for a better understanding.
The smaller circle touches the larger circle at the point $P$. Draw a tangent to the smaller circle at the point $P$. Since the larger circle touches the smaller circle at $P$ as well, the same tangent is also a tangent to the larger circle.
Now, recall:
Therefore, if we draw a perpendicular from the tangent at $P$, it will pass through the center $O$ as well as the center $B$ as it is a tangent to both the circles. Since the perpendicular at $P$ passes through $O$ and $B$ both, it is obvious that the line $PB$ passes through $O$.