In the adjoining figure, $$A$$ is the midpoint of the arc $$BAC$$. Given that $$AB = 15$$ and $$AD = 10$$, the value of $$AE$$ is

We were given that $$A$$ is the midpoint of the arc $$BAC$$ and $$AB=15$$ cm

Since, Arc $$BA=$$ Arc $$AC$$, so the chords $$BA$$ and $$AC$$ are also equal

Hence, we get: $$AC=15$$ cm and $$\angle ACB=\angle ABC$$ as shown below:

Now, Arc $$AC$$ makes two angles, $$\angle ABC$$ and $$\angle CEA$$, in the same segment.

Hence, $$\angle ABC=\angle CEA$$

Also, Arc $$AB$$ makes two angles, $$\angle ACB$$ and $$\angle BEA$$, in the same segment.

Hence, $$\angle ACB=\angle BEA$$

Finally, from the above relationships, we get: $$\angle BE=\angle CEA$$

Or, $$ED$$ is the angle bisector of angle $$E$$ in triangle

Let us assume that $$BD=x$$ cm and $$DC=y$$ cm

So, by the angle bisector theorem:

$$BE=xk$$ cm and $$EC=yk$$ cm , where $$k$$ is a constant

Now, triangle $$BAD$$ and triangle $$ECD$$ are similar to each other

So, $$\dfrac{AB}{EC}=\dfrac{AD}{DC}$$

$$\dfrac{15}{yk}=\dfrac{10}{y}$$

$$k=1.5$$

So, $$BE=1.5x$$ and $$EC=1.5y$$

Now, using Pitot's theorem in cyclic quadrilateral $$ABEC$$

$$AB*CE+AC*BE=BC*AE$$

$$15*1.5y+15*1.5x=BC*AE$$

$$1.5*15(x+y)=(x+y)*AE$$

$$AE=22.5$$ cm

Hence, Option C is correct