Geometry: In the $\triangle ABC, AB=8, BC=7, CA=6$. Let $E$ be a point on $BC$.

Question

In the $\triangle ABC, AB=8, BC=7, CA=6$. Let $E$ be a point on $BC$ such that $\angle BAE=3\angle EAC$. Find $\frac{4(AE)^2}{5}$.

In my solution I had started with the apollonius's theorem which pointed out to me $35.1$ as the answer. There are a few things I don't get -

• Can someone kindly try a solution with the angle relations?
• The answer key says $27$. Where could I have gone wrong?

P.S - I joined $AE$ since it was trivial and assumed it to be the median.

Answer 1

Your conditions for the triangle are inconsistent: $\angle BAE=3\angle EAC$ is not only superfluous but in conflict with the previous conditions. The triangle is already uniquely determined by $AB=8, BC=7, CA=6$ and $E$ being the midpoint of $BC$.

Answer 2

Let $AD$ be the bisector of $\angle BAC$. Then by the angle bisector theorem we get: $$ BD=4,\quad CD=3. $$ The length of bisector $AD$ can be computed by the formula: $$ AD=\sqrt{AB\cdot AC- BD\cdot CD}=6. $$ Hence $ACD$ is an isosceles triangle. The condition $∠=3∠$ means that $AE$ is the bisector of $\angle DAC$ (not the midpoint of $BC$) and it is thus easy to find the solution $AE=\sqrt{135}/2$.

Answer 3

CASE I: Point $E$ is between $B$ and $C$.

Let $AF$ be bisector of $\angle A$ then $\angle BAF=\angle CAF=2\alpha$

$\frac{AB}{AC}=\frac{BF}{CF}$, $\frac{8}{6}=\frac{BF}{7-BF}$, then $BF=4$ and $CF=3$

$AF=\sqrt{AB\times AC - BF \times CF}=\sqrt{8\times 6 - 4\times 3}=6$

and

$\angle FAE=\angle CAE=\alpha$ then $AE\perp BC$

$AE^2 +FE^2=AF^2$ (Pythagorean theorem)

$AE^2=6^2-(\frac{3}{2})^2=\frac{135}{4}$

$\frac{4(AE)^2}{5}=27$

CASE II: Point $E$ is on side $C$

$\angle BAC=2\angle CAE = 2\alpha$

Let $AF$ be bisector of $\angle BA$C then $\angle BAF=\angle CAF=\alpha$

then

$BF=4$, $CF=3$ and $AF=6$

and $AC$ is the bisector of $\angle FAE$ and $\triangle ABF \cong \triangle AEC$ then $AE=8$

$\frac{4(AE)^2}{5}=\frac{256}{5}$