# Construct a triangle, given the altitude, median, and angle bisector for a vertex.

We are given that in a triangle, say $$\triangle ABC$$, the altitude is dropped from A to the opposite side of the triangle. Also given is the median from A and it's the angle bisector.

With help of the above conditions, construct $$\triangle ABC$$.

I tried solving this problem. I know that if it is an equilateral or isosceles triangle the question can be easily done. But there should definitely be a general proof for any triangle.

• The conclusion of your construction statement is incomplete. Construct a triangle satisfying what conditions? Commented Jun 30, 2020 at 13:55
• Are altitude, median, and angle bisector "given" as drawn lines, by their length, or both? Commented Jun 30, 2020 at 14:33
• In this recent answer, I determine triangles based on equal median, altitude, and bisector from three vertices. That's a different problem than this one, of course, but the equations $(1)$, $(2)$, $(3)$ give the lengths of those segments in terms of the triangle sides; "all you have to do" is solve the system for the side-lengths in terms of the segment-lengths. (Well, you first need to be sure to adjust the equations so that they all refer to segments emanating from the same vertex.)
– Blue
Commented Jun 30, 2020 at 14:42
• @JAOFELIX Please avoid writing entire sentences using capital letters. It comes off as quite rude, even if being rude is not your intention. (Some people read it as if the author were shouting.) Commented Jun 30, 2020 at 14:44
• having the height from A to a and the angle bisector at A , is not enough to construct a triangle you get many with tis two the same. Commented Jun 30, 2020 at 20:08

Couple of lemmata proven below helps us to concisely elucidate the proposed construction. Unless stated otherwise, we use the expression “$$\mathrm{angle\space bisector}$$” to denote the $$\mathrm{interior\space angle\space bisector}$$ of an angle. $$\mathbf{Lemma\space 1.1}$$

The altitude and the median dropped from a given vertex of all scalene triangles lie on either side of the interior angle bisector at that vertex.

$$\mathbf{Proof\space 1.1}$$

Consider $$\mathrm{Fig.\space 1}$$, where $$M, D$$, and $$H$$ are the respective feet of the median, the angle bisector, and the altitude dropped from the vertex $$A$$ of an scalene triangle $$ABC$$.

Let $$\measuredangle B \gt \measuredangle C$$. Therefore, $$CA \gt AB$$. We know that, by definition, $$BM = MC = \frac{1}{2}BC$$. We also know that $$DC :BD = CA : AB$$. Therefore, $$DC \gt BD$$, which means that $$DC \gt \frac{1}{2}BC =MC$$. $$\therefore\quad M\space \mathrm{lies\space between}\space D\space \mathrm{and}\space C. \tag{1}$$

Since $$\measuredangle B \gt \measuredangle C$$, we have $$\measuredangle HAB \lt \measuredangle CAH$$. This means that $$\measuredangle HAB \lt \measuredangle DAB = \frac{1}{2}\measuredangle A$$ or $$H$$ lies between $$B$$ and $$D$$. $$\therefore\quad H\space \mathrm{lies\space between}\space B\space \mathrm{and}\space D \tag{2}$$ Statements (1) and (2) together prove Lemma 1.1.

$$\mathbf{Lemma\space 1.2}$$

If feet of any two lines mentioned above coincide, then the foot of the remaining line coincides with the feet of the other two.

$$\mathbf{Proof\space 1.2}$$

For instance, if the foot of the median coincides with that of the angle bisector, we have $$BD = DC \quad\rightarrow\quad \frac{BC\cdot AB}{AB+CA} = \frac{BC\cdot CA}{AB+CA} \quad\rightarrow\quad AB = CA.$$

This proves that $$ABC$$ is an isosceles triangle with its apex at $$A$$. In an isosceles triangle, feet of all three lines mentioned above coincide.

The other cases can be proved using similar arguments.

$$\mathbf{Lemma\space 2}$$

The point of intersection of the extended angle bisector of a given vertex of a scalene triangle and the perpendicular bisector of the opposite side of that vertex lies on the circumcircle of that triangle.

$$\mathbf{Proof\space 2}$$

We consider the angle bisector of the $$\measuredangle A$$ (i.e. $$AE$$) and the perpendicular bisector of the side $$BC$$ shown in $$\mathrm{Fig.\space 2}$$. These two lines meet at $$F$$. Let $$\measuredangle BCA = \phi$$ and $$\measuredangle CAE = EAB = \alpha$$. Then $$\measuredangle CEF$$, which is one of the exterior angles of the triangle $$AEC$$ is equal to $$\left( \alpha + \phi\right)$$. This is also one of the exterior angles of the triangle $$DFE$$. Therefore, $$\measuredangle DFE = \alpha + \phi – 90^o. \tag{3}$$ Let $$O$$ be the circumcenter of the triangle $$ABC$$. Hence, the perpendicular bisector of the side $$BC$$ (i.e. $$DF$$) passes through $$O$$. We can write that $$\measuredangle BOA$$, the angle subtended at $$O$$ by the side $$AB$$, is equal to $$2\phi$$. Since $$OA = OB$$, $$OAB$$ is an isosceles triangle. Therefore, $$\measuredangle OAB$$ is equal to $$90^o - \phi$$, which means that $$\measuredangle EAO = \alpha + \phi – 90^o. \tag{4}$$ Equations (3) and (4) confirm that $$OFA$$ is an isosceles triangle. Therefore, $$OF = OA$$ = Circum-Radius - meaning $$F$$ lies on the circumcircle of $$ABC$$.

Please note that this lemma is not applicable to isosceles and equilateral triangles, because it is not possible to define the point $$F$$.

$$\mathbf{Construction}$$

The construction of the triangle $$ABC$$ is carried out in two separate stages. In the first stage, the line, on which the side $$BC$$ lies, is found after line segments representing the given altitude, angle bisector, and median are laid out in space. In the second stage, the circumcircle of $$ABC$$ is constructed after finding its center and a point that lies on its circumference. The two vertices $$B$$ and $$C$$ are the points of intersection between the circumcircle and the line that contains the side $$BC$$. $$\mathbf{Stage\space 1}$$

We make use of the fact that side $$BC$$, altitude, and angle bisector forms a right triangle to lay out these three lines in space as shown in $$\mathrm{Fig.\space 3}$$. First, a circle having $$AD$$ as its diameter is drawn with its center at $$P$$, which is the midpoint of the angle bisector $$AD$$. A second circle is drawn having the length of the altitude as its radius and $$A$$ as its center. Any one of the two points of intersection between these two circles can be selected as $$H$$, the foot of the altitude. The line $$HD$$ contains the side $$BC$$.

Now, construct another circle having the length of the median as its radius and $$A$$ as the center to cut the extended $$HD$$ at $$M$$ and $$N$$. In accordance with Lemma 1.1, we have to select $$AM$$ as the median. If we select $$AN$$ instead, we are putting altitude and median on the same side of angle bisector. Selection of $$AM$$ as the median define $$M$$ as the midpoint of side $$BC$$.

$$\mathbf{Stage\space 2}$$

Draw the perpendicular line $$MF$$ to $$HD$$ at $$M$$ to intersect the extended angle bisector $$AD$$ at $$F$$ as depicted in $$\mathrm{Fig.\space 4}$$. According to Lemma 2, $$F$$ is located on the circumcircle of the sought triangle $$ABC$$. Therefore, $$AF$$ is a chord of this circumcircle, the center of which lies on $$EQ$$, the perpendicular bisector of $$AF$$. Furthermore, since $$M$$ is the midpoint of side $$BC$$ and $$MF$$ is perpendicular to the side $$BC$$, the circumcenter of $$ABC$$ lies on $$MF$$ as well. This means that the point of intersection of $$EQ$$ and $$MF$$ is the circumcenter $$O$$ of $$ABC$$. Now, to complete the construction, draw the circumcircle, which has the length of $$AO$$ as its radius and $$O$$ as its center to cut the extended $$HD$$ at $$B$$ and $$C$$.

$$\mathbf{Additional\space Information}$$

For brevity, let length of altitude, median, and angle bisector be equal to $$h$$, $$m$$, $$d$$ respectively.

The above described construction produces a unique triangle, if an only if $$m \gt d \gt h \gt 0$$. The case mentioned in Lemma 1.2, i.e. $$m = d = h \gt 0$$, where the sought triangle is either an isosceles or an equilateral triangle, can lead to infinite number of solutions. Collapsing of altitude, median, and angle bisector on to a single line makes this case an underdetermined problem and allows the side $$BC$$ to have any value.

Stage 1 of the construction could have been carried out in two more ways. Firstly, instead of the right triangle already mentioned, we could have constructed the right triangle formed by side $$BC$$, altitude, and median and continued accordingly. Secondly, since both right triangles have altitude as one of their sides, it is also possible to copy one of them on to the other while observing Lemma 1.1. The last method has an advantage over the other two because we do not have anything to exclude.

At the end of the stage 1 of our construction, we have excluded the median $$AN$$ (see $$\mathrm{Fig.\space 3}$$) from our solution space citing a violation of Lemma 1.1. Nevertheless, one can carry out the stage 2 of the construction taking $$AN$$ as the median to obtain a triangle as the solution, if $$h$$, $$m$$, and $$d$$ satisfies the following condition. $$\frac{1}{h^2} \ge \frac{1}{m^2} + \frac{1}{d^2} \tag{5}$$

This triangle turns out to have the same altitude and median as the sought triangle. But, the prescribed length of the angle bisector corresponds to that of the exterior angle bisector. This outcome is possible and correct because Lemma 1.1 is not applicable to the bundle of altitude, median, and exterior angle bisector. If the values of $$h$$, $$m$$, and $$d$$ upholds the equal sign of (5), (e.g. $$h=12$$, $$m=20$$, and $$d=15$$), the resulting triangle is the degenerated triangle with $$BC=0$$.

• thanks 4 ur efforts @YNK. I really appreciate the time u have spent in helping me. :) Commented Aug 9, 2020 at 12:59

Construct a triangle, given three distinct lengths of cevians from the same vertex, say, $$A$$, the median $$m_a$$, the angle bisector $$\beta_a$$ and the altitude $$h_a$$, $$h_a<\beta_a< m_a$$.

Known relations between $$m_a,\, \beta_a,\, h_a$$ and the side lengths $$a,b,c$$ are:

\begin{align} 2m_a^2&=b^2+c^2-\tfrac12a^2 \tag{1}\label{1} ,\\ \beta_a^2&= bc\left(1-\frac{a^2}{(b+c)^2}\right) \tag{2}\label{2} ,\\ h_a^2 &= \frac{4b^2c^2-(b^2+c^2-a^2)^2}{4\,a^2} \tag{3}\label{3} . \end{align}

Excluding $$b,c$$ from \eqref{1}-\eqref{3}, we get a quadratic expression in $$a^2$$

\begin{align} (a^2)^2-8\,(m_a^2-2h_a^2)\,a^2 &+\frac{16(m_a^2-\beta_a^2)(\beta_a^2\,m_a^2-\beta_a^2\,h_a^2-m_a^2\,h_a^2)}{\beta_a^2-h_a^2} =0 \tag{4}\label{4} , \end{align} which gives the value of the side length $$a$$. Equation \eqref{1} provides the value of $$b^2+c^2$$ in terms of $$a$$, and \eqref{3} provides the value of $$b^2c^2$$ in terms of $$a$$, which gives another quadratic equation with the roots $$b^2,c^2$$.

So the expressions for the side length $$a$$ and the other two side lengths in terms of $$a$$ are

\begin{align} a_{1,2}&= 2\sqrt{m_a^2-2h_a^2 \pm (2h_a^2-\beta_a^2)\sqrt{\frac{m_a^2-h_a^2}{\beta_a^2-h_a^2}}} \tag{5}\label{5} ,\\ b,c&= \tfrac12\sqrt{4m_a^2+a^2 \pm 4a\sqrt{m_a^2-h_a^2}} \tag{6}\label{6} . \end{align}

Example

\begin{align} h_a&=3 ,\quad \beta_a=4 ,\quad m_a=5 . \end{align}

Eq. \eqref{4} gives two roots

\begin{align} a_1&= \tfrac27\,\sqrt{343+56\sqrt7} \approx 6.332 ,\\ a_2&= \tfrac27\,\sqrt{343-56\sqrt7} \approx 3.988 , \end{align}

corresponding \begin{align} b_1,c_1&= \tfrac27\,\sqrt{392+14\sqrt7 \pm 14\sqrt{343+56\sqrt7}} \approx 7.768649668. 3.113762020 ,\\ b_2,c_2&= \tfrac27\,\sqrt{392-14\sqrt7 \pm 14\sqrt{343-56\sqrt7}} \approx 6.702893563, 3.608848334 . \end{align}

Verification confirms that triple $$(a_1,b_1,c_1)$$ indeed corresponds to the triangle with given $$h_a,\beta_a,m_a$$:

Edit

As the
answer to the follow-up question proves, the only valid root is always $$a_1$$,

\begin{align} a_{1}&= 2\sqrt{m_a^2-2h_a^2 + (2h_a^2-\beta_a^2)\sqrt{\frac{m_a^2-h_a^2}{\beta_a^2-h_a^2}}} . \end{align}

• @gcov It is good that you did some follow-up to your own answer. When I drew the triangle using $a_2, b_2,$, and $c_2$, I found that the angle bisector is longer than 4. However, Aretino’s answer to your question does not shed any light on why you get this second real solution in the first place. Using the eaqution (5) of your answer I found that, If $$\frac{1}{h^2}\ge\frac{1}{m^2}+\frac{1}{\beta^2},$$ there is always two real solutions. Do you have any idea why? The second reason why I am writing this comment is given in my next comment. $$\mathrm{Contd.}$$
– YNK
Commented Jul 4, 2020 at 17:08
• @gcov I have already found a geometric construction, which uses straight edge and compasses, to solve this problem. I also get 2 different solutions if the mentioned condition is true. I cannot find a reason to reject this $2^{nd}$ solution without first drawing the corresponding angle bisector. I would like to show this to you, but, unfortunately, we have been barred from answering this question. Is there anything you (you have more privileges than I do) can do to re-open the question to receive answers? Please read my yesterday's comment and a reply to it by Blue.
– YNK
Commented Jul 4, 2020 at 17:12
• @YNK: I've already voted to reopen the question, there are 4 votes at the moment, just one more needed. Need to look closely at your comment on $a_2$. Can you compose your these comments into the answer to the follow-up question? Commented Jul 4, 2020 at 17:36
• @YNK: And btw, there is a perfectly accepted way not to waste your precious answer in case the question is closed. You can ask an improved self-answered question. Commented Jul 4, 2020 at 17:51
• @gcov I have re-posted my comments as you requested. Thank you for your advice and information.
– YNK
Commented Jul 4, 2020 at 17:56

In the indicated case it is given that the median $$m_a$$ from $$A$$ and the angle bisector $$w_A$$ at $$A$$ coincide. In this case the triangle is isosceles, and the altitude from $$A$$ coincides with the line $$m_a=w_A$$. It follows that the three occurring lengths coincide, and are equal to the height $$h_a=|AM_a|$$, where $$M_a$$ is the midpoint of $$BC$$ on the base of the triangle. When the given lengths $$m_a$$, $$\beta_a$$, and $$h_a$$ do not coincide there is no triangle having the desired properties.

On the other hand, when $$m_a=\beta_a=h_a$$, the given information does not determine the two points $$B$$, $$C$$ at equal distance from $$M_a$$ on the base.

• @Chritian Blater thanks for the effort you have put but in the question we are never given that the length of angle bisector, median and altitude is equal. We have to find something in general. Commented Jun 30, 2020 at 15:44