# 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?

• Here's a late response, using synthetic geometry; I only recently noticed this interesting post. May 8, 2018 at 16:48

It can also be proved using geometry only.

1] Through $$M$$, draw $$MX \parallel BC$$ cutting $$AC$$ at $$X$$. From that, we have

• (1.1) $$\beta = \beta’$$; and
• (1.2) $$AX = XC$$.

2] Through $$X$$, draw $$XYZ \parallel AB$$ 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 $$\alpha’ = \alpha = \beta = \beta’$$. 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.

• Could you explain why $CM$ is the diameter of the dotted circle? Why is $\angle HXC$ a right angle? Why is $\angle ACB$ a right angle?
– robjohn
Jan 5, 2021 at 22:32
• @robjohn CH is the given as the altitude of $\triangle ABC$. This means $\angle CHM = 90^0$. By converse of angle in semi-circle, CM must be the diameter of the circle passing through CHM.
– Mick
Jan 6, 2021 at 12:46
• Okay, I agree that $CM$ is the diameter of the circle (if this explanation were included in your answer, it would be a lot clearer), but it would seem that $\angle MXC$ would be a right angle, not $\angle HXC$ (it would help if $X$ were actually shown on the circle). It would be helpful to mention the role that the angle between $\alpha$ and $\beta$ plays.
– robjohn
Jan 6, 2021 at 13:21
• @robjohn The point that I want to make is ‘initially we don’t know whether X is one of the con-cyclic point of the dotted circle’. That is why it was intentionally not drawn on the circle. You are right about the typo of $\angle HXC$. Fixed.
– Mick
Jan 6, 2021 at 16:57
• (+1) I have added a section to my answer clarifying what seemed confusing to me.
– robjohn
Jan 7, 2021 at 21:30

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.

Trigonometric Approach

We are given that $$\angle MAC=\angle BAP$$ and since $$\triangle BPA$$ is a right triangle, $$\angle BAP=\frac\pi2-B$$. Therefore, $$\angle MAC=\frac\pi2-B\tag1$$ and, since $$\angle MAC+\angle BAM=A$$, we have $$\angle BAM=A+B-\frac\pi2\tag2$$ Since $$M$$ is the bisector of $$\overline{BC}$$, the areas of $$\triangle BAM$$ and $$\triangle MAC$$ are equal. Thus, \begin{align} \overbrace{\frac12cm\sin\left(A+B-\frac\pi2\right)}^{\left|\triangle BAM\right|} &=\overbrace{\frac12bm\sin\left(\frac\pi2-B\right)}^{\left|\triangle MAC\right|}\tag{3a}\\ c\cos(C)&=b\cos(B)\tag{3b}\\[3pt] c\sin(B)&=b\sin(C)\tag{3c}\\[3pt] \sin(2B)&=\sin(2C)\tag{3d} \end{align} Explanation:
$$\text{(3b)}$$: apply $$A+B-\frac\pi2=\frac\pi2-C$$ and multiply by $$\frac2m$$
$$\text{(3c)}$$: Law of Sines
$$\text{(3d)}$$: cross multiply $$\text{(3b)}$$ and$$\text{(3c)}$$ and multiply by $$\frac2{bc}$$

Since $$\triangle ABC$$ is not isosceles, $$\text{(3d)}$$ implies that $$2B=\pi-2C$$; which, since $$A=\pi-B-C$$, means $$\bbox[5px,border:2px solid #C0A000]{A=\frac\pi2}\tag4$$

Geometric Approach

This is an approach similar, but slightly different, to that in Mick's answer.

Start with the diagram above. We are given that $$\angle MAC$$ is equal to $$\angle BAP$$, which is $$\frac\pi2-B$$ since $$\angle APB$$ is a right angle.

Add $$N$$ and $$R$$, the midpoints of $$\overline{AC}$$ and $$\overline{AB}$$ respectively. Then add the segments from $$R$$ to $$N$$, $$M$$, and $$P$$.

Add the circle with $$\overline{AM}$$ as its diameter. Since $$\angle APM$$ is a right angle, $$P$$ sits on this circle. Since $$\overline{RN}$$ is parallel to $$\overline{BC}$$, $$\overline{RN}$$ is also perpendicular to $$\overline{AP}$$.

Since $$\triangle ARN$$ is similar to $$\triangle ABC$$ with a linear ratio of $$\frac12$$, $$Q$$, the intersection of $$\overline{AP}$$ and $$\overline{RN}$$, bisects $$\overline{AP}$$. Thus, $$\angle RPQ$$ is also $$\frac\pi2-B$$.

Since $$\overline{RM}$$ is parallel to $$\overline{AC}$$, $$\angle RMA$$ is equal to $$\angle MAC$$, which equals $$\frac\pi2-B$$.

The locus of points at which $$\overline{AR}$$ subtends an angle of $$\frac\pi2-B$$ is the circle which contains $$A$$, $$R$$, $$P$$, and $$M$$. This is the same circle that was added above since it contains $$A$$, $$P$$, and $$M$$. Thus, $$\angle ARM$$ is a right angle, and since $$\overline{RM}$$ is parallel to $$\overline{AC}$$, so is $$A$$.

• Great answer! Did you keep this question in your backlog and then came back to it? It's been a while. Jan 6, 2021 at 20:16
• Thanks! Actually, it showed up on the main list. Once I realized how old the question was, I looked for the edit that put it onto the main list. It turns out it was an automatic bump by the system.
– robjohn
Jan 6, 2021 at 21:35

Extend median $$\overline{AM}$$ of $$\triangle ABC$$ to meet the circumcircle at $$M'$$, and let $$D$$ be the foot of the altitude from $$A$$. (Note that $$D$$ and $$M$$ are distinct —and, thus, $$\overline{AM'}\not\perp\overline{BC}$$— because we assume $$\triangle ABC$$ is non-isosceles.)

Since $$\angle M' \cong\angle B$$ (by the Inscribed Angle Theorem), we conclude $$\angle ACM'\cong\angle ADB=90^\circ$$, so that $$\overline{AM'}$$ is a diameter. Now, a diameter can only bisect a non-perpendicular chord at its own midpoint (aka, the circumcenter); consequently, $$M$$ is that circumcenter, $$\overline{BC}$$ is a diameter, and $$\angle BAC$$ is a right angle. $$\square$$

• Amazing synthetic answer. Thanks for your contribution! Jan 10, 2021 at 16:29