The problem and answer are from a book.

Is there a $\triangle ABC$ such that the altitude from $A$, the bisector of $\angle BAC$ and the median from $A$ divide $\angle BAC$ into four equal parts?

The answer is: $ABC$ is a right triangle with $\angle A=90^\circ$ and $\angle C=22.5^\circ$.

I don't know how I should start to solve this problem.

  • 1
    $\begingroup$ Do you mean altitude from A bisects $\angle BAC$? if so please edit your question. $\endgroup$
    – sirous
    Jan 2, 2020 at 14:58
  • $\begingroup$ @sirous Yes Sir I will modified $\endgroup$ Jan 2, 2020 at 15:15

3 Answers 3


This is not difficult.

Let the altitude be $AE$, the bisector $AD$ and the median $AM$.

Let $\angle A=4\alpha$. Since $\angle BEA=90^\circ$, we have $\angle B=90-\alpha$. Hence $\angle C=90-3\alpha$ (because the angles in $ABC$ add to $180^\circ$).

enter image description here

From triangle $MAC$ the sine rule gives $MC/MA=\sin\alpha/\sin(90^\circ-3\alpha)=\sin\alpha/\cos3\alpha$. Similarly, from triangle $MAB$ we get $MB/MA=\sin3\alpha/\cos\alpha$. But $MB=MC$, so $\sin3\alpha\cos3\alpha=\sin\alpha\cos\alpha$. Hence $\sin6\alpha=\sin2\alpha$, so $6\alpha+2\alpha=180^\circ$ and hence $\alpha=22.5^\circ$ and so $\angle BAC=90^\circ$.

  • $\begingroup$ Nice way sir , Thank you very much!😍😍 $\endgroup$ Jan 2, 2020 at 18:35

Well the angle bisector cuts $\angle BAC$ into two equal parts, so the median and the altitude must each cut those two angles into equal parts. If we label point where the altitude intersects $BC$ as $J$, the point where the angle bisector as $K$, and the median as $M$, we know that $K$ is colinearly between $J$ and $M$. As it is arbitrary which endpoint is $B$ and which is $C$ we can assume that points lie in order as $B,J,K,M$

So we have a figure a big triangle $\triangle ABC$ divided into four smaller triangles.

$\angle BAJ \cong \angle JAK \cong \angle KAM \cong \angle MAC$.

$\angle BJA \cong \angle KJA$ are both right angles.

So $\triangle BJA \cong \triangle KJA$.

If we let $m\angle BAJ = m\angle JAK = m\angle KAM = m \angle MAC= X$ then we can conclude:

$m\angle ABJ = 90 -X$ and $m\angle BCA = 180-(90-X)-4x = 90-3X$.

Now if we look at the line $BC$ and use trig identities we know.

$\frac {BJ}{AJ} = \tan X$.

$\frac {KJ}{AJ} = \tan X$. (And $BJ=JK$)$

$\frac {MJ}{AJ} = \tan 2X$.

And $\frac {CJ}{AJ} = \tan 3X$.

And heres the lynchpin: $M$ is the midpoint of $BC$ so $MC = MB$.

Now $MC = CJ - MJ = AJ(\tan 3X - \tan 2X)$ and $MB = BJ + MJ = AJ(\tan X + \tan 2X)$.

So we have $\tan 3X - \tan 2X = \tan X + \tan 2X$

  • $\begingroup$ its a perfect solution ! $\endgroup$ Jan 2, 2020 at 18:37
  • $\begingroup$ Not that perfect. I didn't want to actually do the work. $\endgroup$
    – fleablood
    Jan 2, 2020 at 19:20

Here is a possible solution which is not using trigonometric functions, but rather only similarities and the angle bisector theorem (ABT).

We denote by $H,J,M$ the points on $BC$ which are respectively on the height, angle bisector, and median of $A$. Let $K$ be the reflection of $J$ w.r.t. the median $AM$. It lies on the side $AC$. Let $a,b,c$ be the side lengths as usual. Let the angle in $A$ have the measure $4x$. Then we can quickly chase the following angles in the picture:

Triangle where the height, angle bisector, median from the vertex <span class=$A$ divide the angle in $A$ in four equal pieces">

Now we compute the lenghts in the picture.

  • From the ABT in $\Delta AJC$ we get the length of $JM$ as shown in the picture, $\frac a2\cdot \frac cb$. So we also have $MK$.
  • $BJ=BM-JM$ leads to $BJ=\frac a2\cdot \frac {b-c}b$.
  • The rest of the segment $BC=a$ is then $JC=\frac a2\cdot {b+c}b$.

We write now the ABT in the triangle $\Delta ABC$, and the similarity $\Delta ABC\sim \Delta MKC$ to get the relations: $$ \begin{aligned} \frac cb=\frac{b-c}{b+c}\ ,\\ \frac ab=\frac{b-c}{a/2}\ . \end{aligned} $$ The two realtions are $b^2-bc=c^2+bc$ and $a^2=2b^2-2bc$. It follows immediately $a^2=b^2+(b^2-2bc)=b^2+c^2$, so the angle in $A$ is $4x=90^\circ$, and all angles can be explicitly identified.

To be pedant, we have to check if the obtained triangle satisfies the condition, yes, this is the case, the height $AH$ is the angle bisector of $\widehat{BAJ}$, and because of $MA=MC$ we have $\widehat{MAC}=\widehat{MCA}=22.5^\circ$, so $AM$ bisects $\widehat{JAC}$.

  • $\begingroup$ why did you decide that you "have" MK? $\endgroup$
    – Mike
    Apr 12 at 20:34
  • $\begingroup$ It is a long time since the solution was typed, very often i am just starting to type without having the full (shortest) path to the final step. "Having" $MK$ is not bad, since any detail can be useful on the road, in this case "knowing $MK$" does not additionally help, since the similarity of $\Delta ABC$ and $\Delta MCK$ leads to $$\frac{CM}{CA}=\frac{CK}{CB}=\frac{MK}{AB}\ ,\text{ i.e.}\\\frac{a/2}b = \frac{b-c}a = \frac{\frac a2\cdot \frac cb}c\ ,$$and the last term is never used, it repeats the first one. (However, it did not hurt to mention it while still searching for a solution.) $\endgroup$
    – dan_fulea
    Apr 12 at 23:04

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