In $\triangle ABC$, if the altitude, median, and angle bisector at $A$ quadrisect $\angle A$, then $\angle A=90^\circ$ and $\angle C=22.5^\circ$

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.

• Do you mean altitude from A bisects $\angle BAC$? if so please edit your question. – sirous Jan 2 '20 at 14:58
• @sirous Yes Sir I will modified – Ellen Ellen Jan 2 '20 at 15:15

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 $$180^\circ$$).

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 $$\angle BAC=90^\circ$$.

• Nice way sir , Thank you very much!😍😍 – Ellen Ellen Jan 2 '20 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 $$\tan 3X - \tan 2X = \tan X + \tan 2X$$ • its a perfect solution ! – Ellen Ellen Jan 2 '20 at 18:37 • Not that perfect. I didn't want to actually do the work. – fleablood Jan 2 '20 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:$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 $$\widehat{JAC}$$.