Problem statement:
Prove that if the altitude and median drawn from the same vertex of a nonisosceles triangle lie inside the triangle and form equal angles with its sides, then this is a right triangle.
After many attempts, I came up with this: let $ABC$ be the triangle where $CH$ is the altitude, $CM$ is the median, and name the angles $ACH = MCB = \theta$, $CAH = \alpha$, $CBM = \beta$ and $HCM = \gamma$. Using the sine theorem in the triangle $CMB$ we get $$\frac{MB}{\sin(\theta)} = \frac{MC}{\sin(\beta)},$$ while using the sine theorem in the triangle $ACM$ we get $$\frac{MC}{\sin(\alpha)} = \frac{MA}{\sin(\theta + \gamma)}.$$ Combining these equations I found $$\sin(\alpha) \sin(\theta) = \sin(\beta) \sin(\theta + \gamma).$$ Applying the identity $$\sin(a)\sin(b) = \frac{\cos(a-b) – \cos(a+b)}{2}$$ I concluded $$\cos(\alpha – \theta) = \cos(\theta + \gamma – \beta).$$ Since all angles are within $[0, \pi]$ range I concluded $$\alpha – \theta = \theta + \gamma – \beta,$$ thus $2 \theta + \gamma = \alpha + \beta$. From the original triangle we know $$2 \theta + \gamma + \alpha + \beta = \pi,$$ and combining the equations proves the triangle is right.
Is this correct? Is there a synthetic way to do it?
Best Answer
It can also be proved using geometry only.
1] Through $M$, draw $MX \parallel BC$ cutting $AC$ at $X$. From that, we have
2] Through $X$, draw $XYZ \parallel AB$ cutting $CH$ at $Y$ and $CM$ at $Z$. From that, we have
(2.1) + (1.1) implies $\alpha’ = \alpha = \beta = \beta’$. Since $H$ and $M$ are distinct ($\triangle ABC$ is not isosceles), the inscribed angle theorem means $XHMC$ is a cyclic quadrilateral. This also gives us that $\angle MXC = \angle MHC = 90^\circ$.
Then since $MX \parallel BC$, also $\angle ACB = 90^\circ$ as required.