geometrytriangles
In an acute angled triangle $ABC$, the angle bisector $AL$, altitude $BH$ and perpendicular bisector of $AB$ are concurrent. What is the $\angle BAC$?
I presume that $\triangle ABC$ has to somehow be an equilateral triangle as I've tried sketching the problem out (albeit highly inaccurately) and found that it is most close to being concurrent when $\triangle ABC$ is equilateral. If this is so, how can we prove it is equilateral using only the information from the question?
Let us consider the triangle $ABC$. Since the triangle is acute, the altitude from $B$ necessarily intersects $AC$ internally, i.e. $E$ is between $A$ and $C$.
Let the lengths and angles of the triangle be labelled according to the figure above.
From the angular bisector theorem, we have $$\frac{c}{b} = \frac{m}{n}$$ where $m = |BD|$ and $n = |CD|$. Likewise, from the generalized angular bisector theorem, we have $$\frac{h\sin\beta}{k\sin\alpha} = \frac{m}{n}$$ where $h$ is the length of the altitude from $B$ and $k=|CE|$. Note that the claim $\alpha > 45^\circ$ is equivalent to $\alpha > \beta$ since $\alpha + \beta = 90^\circ$. Therefore the claim is true if and only if $\alpha > \beta$ if and only if $\sin\alpha > \sin\beta$. Therefore the problem is equivalent to proving $$\frac{c}{b} = \frac{h\sin\beta}{k\sin\alpha} < \frac{h}{k}$$ Let $hb = \Delta$ denote the area of the triangle. Then equivalently $$k < \frac{\Delta}{c} = h_c$$ where $h_c$ is the length of the altitude from $c$. The above inequality is easily seen to be true. Consider swinging the length $CE$ around $C$ to form a circle of radius $k$ centered at $C$. Then the line $BE$ is tangent to the circle at $E$ and clearly separates the circle and the side $AC$. The claim follows.
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 $BC=0$.
Best Answer
As in the above drawing let $\angle DAL=\angle OAC=a$
And let be the intersection of the angle bisector, perpendicular bisector and the altitude be $O$
Let the perpendicular bisector be $DK$
$\angle DAO=\angle OAH=a and \angle AOH=\angle DOA=90-a$
Then $\triangle DAO \equiv \triangle OAH (A.A.S)$
With this you can get that $AD=AH$ and $OD=OH$
Since $DK$ is the perpendicular bisector of $AB$ $AD=DB$1
Now draw $RH$ and also the median respective to the right angle in a right angles triangle is equal to the half of the hypotenuse
$AD=AH$
And that brings us to $AD=AH=DH$
Since $\triangle ADH$ is equilateral
$\angle BAC=\angle ADH=\angle DHA=60$