Question

In $△ABC$, the median $AM\perp AB$. Evaluate $\left|\frac{ab}{c^2}\cos A\right|$

Answer 1

Length of median $AM=m=\frac{1}{2}\sqrt{2b^2+2c^2-a^2}$

Applying pythagoras theorem in $△ABM$, we have

$\frac{a^2}{4}=c^2+m^2$

$⟹a^2=b^2+3c^2$

Also, from the cosine rule,

$cosA=\frac{b^2+c^2-a^2}{2bc}$

$⟹cosA=-\frac{c}{b}$

$\therefore \left|\frac{ab}{c^2}cosA\right|=|-\frac{a}{c}|=\frac{a}{c}$