Question

In a triangle ABC, $a=7,b=8,c=9$, BD is the median and BE the altitude from the vertex B, then

Answer 1

In $△ABC,cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{8^2+9^2-7^2}{\mathrm{2.8.9}}=\frac{2}{3}$
And in $△ABD,$ $cosA=\frac{{AB}^2+{AD}^2-{BD}^2}{2.AB.AD}=\frac{9^2+4^2-{BD}^2}{\mathrm{2.9.4}}$
$\frac{97-{BD}^2}{72}=\frac{2}{3}\Rightarrow {BD}^2=49\Rightarrow BD=7$
$\Rightarrow △BCD$ is isosceles triangle
$\Rightarrow CE=ED=2$
Now $△BED$ ${BD}^2={BE}^2+{ED}^2$
$\Rightarrow BE=\sqrt{7^2-2^2}=\sqrt{45}$
And now $AE=AD+ED=4+2=6$.