Question

In the given figure, BC is a tangent to the circle with centre O. OE bisects AP. Prove that $△AEO\sim △ABC.$

Answer 1

In the fig $\left(a\right)$

$BC$ is the tangent to the circle with centre $O$

$OE$ bisect $AP$

also given to prove that

$△AEO\sim △ABC$

Now, from the fig $\left(a\right)$

Let is consider

$△AEO$ & $△PEO$

So we get

$OE=OE$ (given)

$EP=EA$ (given)

$OA=OP$ ($because$ radius of the circle is equal)

$\Rightarrow △AEO\cong △PEO$

$\Rightarrow \angle AEO=\angle PEO={90}^o\Rightarrow$ [ $\because$ As per the property any perpendicular down from the centre of the circle to its chord bisects the circle]

So from fig if we consider $△AEO$ & $△PEO$

So, there by we get,

$|angleAEO=\angle PEO={90}^o$

Now, let us consider $△AEO$ and $△ABC$

we get,

$\angle A=\angle A$ (common angle)

$\angle AED=\angle ABC={90}^o$

As according to the property any perpendicular drawn

from the centre of the circle to its chord bisect the circle

there by

$\Rightarrow △AEO\sim △ABC$

[$\because$ by $AA$ similarity]

$\therefore △AEO\sim △ABC$

Hence proved