# Find the ratio in which the circle divides each of the sides AB and AC?

A circle passes through the vertex A of an equilateral triangle ABC and is tangent to BC at its midpoint .

Find the ratio in which the circle divides each of the sides AB and AC?

Does the line joining the vertex A and the midpoint of BC will be perpendicular to BC? How can i achieve this? Thanks in advance.

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Let us assume that the point where $AC$ intersects the circle is $D$ such that $AD = x, DC=y$

If we drop a line perpendicular from $A$ to $BC$, meeting $BC$ at $P$ and call that height $h$.

$$h=2r = \frac{\sqrt{3}a}{2} \Rightarrow r = \frac{\sqrt{3}a}{4}$$

Also $CP=\frac{a}{2}$ and if we assume $AD=x, DC = y$, then

$$(x+y) = a, \hspace{5pt} CP^2 = DC . AC = y(x+y) = ay$$

$$\frac{a^2}{4} = ay \Rightarrow \left( y = \frac{a}{4}, x=\frac{3a}{4} \right)$$

Therefore the ratio $$x:y = 3:1$$

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Let $|AB|=|AC|=x+y ~\text { and } ~ x > y$, then :

$x=2r \cdot \sin 60^{\circ}$

$y=a-2r \cdot \sin 60^{\circ}$

$r=\frac{a\sqrt 3}{4}$

Hence :

$x=\frac{3}{4} a \text { and } y=\frac{1}{4}a$

therefore :

$x : y =3 : 1$

Answer to your second question is affirmative .

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