A triangle has side length $13,14,15$, and its circumcircle is constructed. The median is then drawn with its base having a length of $14$, and is extended to the circle. Find its length.

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    – Guy
    Mar 25 '14 at 20:27

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Here, $AB=13, AC=15, BC=14$. $AD$ is the median on $BC$ and we have to find the length of $AE$. So, we first take out the length of $m=AD$. Noting by law of cosines that:

$$m^2=c^2-(\frac{a}2)^2+am\cos \alpha$$ $$m^2=b^2-(\frac{a}2)^2+am\cos \beta=b^2-(\frac{a}2)^2-am\cos \alpha$$

And adding them up, to get the Apollonius formula: $$2m^2=c^2+b^2-\frac{a^2}2$$

Thus we solve for $m$, and then substitute the values: $$m=\frac{\sqrt{2c^2+2b^2-a^2}}2=\frac{\sqrt{2\cdot13^2+2\cdot15^2-14^2}}2=2\sqrt{37}$$

Now, we will use the fact that:

$$AD\cdot DE=BD\cdot DC$$ $$2\sqrt{37}\cdot DE=49$$


$$AE=AD+DE=\frac{49}{2\sqrt{37}}+2\sqrt{37}=\frac{197}{2\sqrt{37}}\approx 16.193$$


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