# In an acute angled triangle $ABC$, the angle bisector $AL$, altitude $BH$ and perpendicular bisector of $AB$ are concurrent. What is the $\angle BAC$?

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?

• Say perp from $C$ to $AB$ meet at $H$. If half angle of $\angle A$ is $\theta$ then $\angle ACH = \theta$ or $\angle CAH = 90^\circ - \theta$ but also $\angle CAH = 2 \theta$. So you get $3 \theta = 90^0$ and so $\angle A = 2 \theta = 60^\circ$. No it does not have to be an equilateral triangle as you can move point $B$ anywhere on line through segment $AB$ without violating any of the above conditions. Sep 27, 2021 at 11:30
• @MathLover Beat me to it. Minor quibbles. Perpendicular bisector of AB is not perp from C to AB (i.e. not necessarily pass through C, which is indirect consequence of point that you made). Also, variable H is overloaded. Assume concurrent point is M. Then $\triangle ABM$ is isosceles. Assuming that $\angle CAB = 2\theta$, then you have that $\angle ABH = \angle ABM = \angle BAM = \theta$. Then, as you indicated, $3\theta = 90^{\circ}.$ Sep 27, 2021 at 11:41
• The following triangles satisfy the concurrency condition: $\triangle ABC$ is $60^\circ:30^\circ:90^\circ$, and $\triangle ABC$ is $60^\circ:90^\circ:30^\circ$. While they are NOT acute triangles, they have something in common with equilateral $\triangle ABC$. Sep 27, 2021 at 11:42
• @peterwhy every triangle with $\angle A = 60^\circ$ satisfies. Sep 27, 2021 at 11:45
• @user2661923 I swapped points $B$ and $C$ in my comments... too late to edit Sep 27, 2021 at 11:48

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

$$\angle BAC=\angle ADH=\angle DHA=60$$