# Prove that if the altitude and median of a triangle form equal angles with sides then the triangle is right.

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?

• your answer is correct. – DeepSea Feb 15 '16 at 7:31
• Here's a late response, using synthetic geometry; I only recently noticed this interesting post. – Edward Porcella May 8 '18 at 16:48

It can also be proved using geometry only.

1] Through $M$, draw MX // BC cutting AC at X. From that, we have (1.1) $\beta = \beta’$; and (1.2) $AX = XC$.

2] Through $X$, draw XYZ // AB cutting CH at Y and CM at Z. From that, we have (2.1) $XY$ is the perpendicular bisector of CH; and (2.2) Z is the midpoint of CM.

(2.1) + (1.1) implies $\alpha’ = \alpha = \beta = \beta’$. This further means XHMC is a cyclic quadrilateral with CM as a diameter.

Result follows from the fact that $\angle HXC = 90^0$.

Let $\triangle ABC$ be a right triangle in the semi-circle with diameter $AB$. Then with altitude $CD$ and median $CM$, since $\triangle CMB$ is isosceles$$\angle \beta=\angle \gamma$$But since by Euclid [Elements, VI, 8]$$\triangle ACD\sim\triangle CBD$$with$$\angle\alpha=\angle\gamma$$therefore$$\angle\alpha=\angle\beta$$Hence in a right triangle, the altitude and median from the vertex of the right angle make equal angles with the adjacent sides.

Now we must prove the converse: If the altitude and median from the vertex of a triangle make equal angles with the adjacent sides, the angle at the vertex is right.

First let the angle at $C$ be obtuse, and suppose, in the figure below, that altitude $CD$ and median $CM$ make $$\angle\alpha=\angle\beta$$

Construct a semi-circle on $AB$ as diameter, extend $AC$ to $E$ on the circumference, join $EB$, $EM$, and drop altitude $EF$.

Then, as shown above, since $\triangle ABE$ is right$$\angle\delta=\angle\epsilon$$And since$$EF\parallel CD$$then$$\angle\delta=\angle\alpha$$and it follows that$$\angle\epsilon=\angle\beta$$making$$\triangle EGB\sim\triangle CGM$$and $EBMC$ therefore a cyclic quadrilateral.

But since $\angle CEB$ is right, $CEBM$ is cyclic only if opposite $\angle CMB$ is also right, that is if $D$ coincides with $M$, making $\triangle ACB$ isosceles, as in the figure below.

But this contradicts the given condition that the triangle is not isosceles.

Therefore, (in the second figure) $EBMC$ is not a cyclic quadrilateral, $\triangle EGB$ and $\triangle CGM$ are not similar, and$$\angle\epsilon\ne\angle\beta$$

Therefore, in obtuse $\triangle ACB$$\angle\alpha\ne\angle\beta$$ And by the same reductio argument it can be shown that$\angle\alpha\ne\angle\beta$when the angle at$C\$ is acute.

If the altitude and median from the vertex of a triangle make equal angles with the adjacent sides, the angle at the vertex is right.