Question

An altitude and a median drawn from the same vertex of a triangle divide the angle at that vertex into .three equal parts. Prove that the angles of that triangle are equal to ${30}^{\circ },{60}^{\circ },and{90}^{\circ }$.

Answer 1

Let $ABC$ be a triangle $AH$ the height from $A$ to $BC$ and $AM$ the median ($M$ the midpoint between $B$ and $C$) without loss of generality let's say that $H$ is between $B$ and $M$ so we have $\Delta BAH$ and $\Delta MAH$ congruent as angles $BHA$ and $MHA$ are equal angle $BAH$ and $MAH$ are equal (per hypothesis of problem) and they share side $AH$. Therefore $BM=HM(orHM=\frac{1}{2}BM)$

So, Lets' see $\Delta AHC$. $AM$ is bisector therefore we have ratios $\frac{MH}{MC}=\frac{AH}{AC}$ but $MH=\frac{1}{2}MB=\frac{1}{2}MC$

therefore $\frac{MH}{MC}=\frac{1}{2}$

So $AHC$ is a right angled triangle at $H$ and $AC=2AH$. So, $sinC=\frac{1}{2}$ and $C={30}^0$

$\angle HAC={60}^0$ and $\angle HAB=\angle HAM=\frac{1}{2}\angle HAC={30}^0$ So $\angle BAC={90}^0$

So, $\angle BAD={60}^0$

$\angle BAC=3\times {30}^0={90}^0$.