Using the cosine law: Given a $\triangle ABC$ with $AB=3$, $AC=8$, and median $AD=5$, find $BC$.

Question

I'm trying to solve the following question but somehow got stuck with no clue how to continue:

Given a $\triangle ABC$ with $$AB=3, AD=5, AC=8$$ where $AD$ is the median of $BC$ (meaning $BD=DC$). Find $BC$.

I've used cosine law and reached that: $$ BD=\sqrt{34-30\cdot \cos y}$$ $$CD= \sqrt{89-80\cdot \cos a}$$ Not sure that I can assume that angle $a=y$ ($\angle DAB=\angle DAC$).

Any help would be appreciated.

Thanks!

Answer 1

Apply the cosine rule as follows instead $$AB^2= AD^2+ BD^2 -2AD\cdot BD \cos \theta $$ $$AC^2= AD^2+ CD^2 - 2AD\cdot CD \cos (\pi-\theta )$$ Substitute $BD=CD=\frac12BC$ and eliminate the variable $\theta$ to obtain $BC$.

Answer 2

By Applonius Theorem suggested by @AlbusDumbledore, \begin{align} AB^2 + AC^2 &= 2(AD^2+BD^2) \\ 3^2 + 8^2 &= 2(5^2 + BD^2) \\ BD &= \frac12\sqrt{3^2 + 8^2 - 2\cdot 5^2} = \frac{\sqrt{23}}{2} \\ BC &= 2BD = \sqrt{23} \end{align}

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Not sure that I can assume that angle $a=y$ ($\angle DAB=\angle DAC$).

No, $AD$ is median, not angle bisector.