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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?

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    $\begingroup$ your answer is correct. $\endgroup$
    – DeepSea
    Feb 15, 2016 at 7:31
  • $\begingroup$ Here's a late response, using synthetic geometry; I only recently noticed this interesting post. $\endgroup$ May 8, 2018 at 16:48

4 Answers 4

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It can also be proved using geometry only.

enter image description here

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.

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  • $\begingroup$ 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? $\endgroup$
    – robjohn
    Jan 5, 2021 at 22:32
  • $\begingroup$ @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. $\endgroup$
    – Mick
    Jan 6, 2021 at 12:46
  • $\begingroup$ 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. $\endgroup$
    – robjohn
    Jan 6, 2021 at 13:21
  • $\begingroup$ @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. $\endgroup$
    – Mick
    Jan 6, 2021 at 16:57
  • $\begingroup$ (+1) I have added a section to my answer clarifying what seemed confusing to me. $\endgroup$
    – robjohn
    Jan 7, 2021 at 21:30
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equal angles of median and altitude in right triangle 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.

The proof is by contradiction.

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 $EF$. unequal angles of median and altitude in obtuse triangle

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. when obtuse triangle is isosceles

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.

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Trigonometric Approach

enter image description here

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 $P$.

enter image description here

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 $A$.

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  • $\begingroup$ Great answer! Did you keep this question in your backlog and then came back to it? It's been a while. $\endgroup$ Jan 6, 2021 at 20:16
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    $\begingroup$ 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. $\endgroup$
    – robjohn
    Jan 6, 2021 at 21:35
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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.)

enter image description here

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. $\square$

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    $\begingroup$ Amazing synthetic answer. Thanks for your contribution! $\endgroup$ Jan 10, 2021 at 16:29

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