BdMO Online Forum

Problem 3:
Triangle $ABC$ is acute with the property that the bisector of $\angle BAC$ and the altitude from $B$ to side $AC$ and the perpendicular bisector of $AB$ intersect at one point. Determine the angle $\angle BAC$.

It's an equilateral triangle.Therefore,ans is 60
We know that in an equilateral triangle, the bi-sectors of angles,altitudes and bi-sectors on sides from opposite points,are same(it can be proved easily). Here orthocenter,centroid,circumcenter also same.so...

How do you prove it equilateral.Please try with a simple case.

sorry to not writing the proof before.
i misunderstood the problem and that's why i took time to prove.but it's easy...
in triangle ABC,suppose bisector of angle A is AD.altitude B to AC is BE.perpendicular bisector of AB is CF.CF,AD,BE intersect in a point.so we can apply ceva's theorem.
\[\frac{AE}{CE}.\frac{CD}{BD}.\frac{BF}{AF}=1\]
OR,\[\frac{AE}{CE}.\frac{CD}{BD}=1[BF=AF]\]
OR,\[\frac{AE}{CE}=\frac{BD}{CD}\]
so DE is parallel to AB
in triangles CBF and CAF ,
\[\angle BFC=\angle AFC\]
\[BF=AF\]
CF is common.then triangle CBF and CAF are sorbosomo.
AC=BC
\[\angle A=\angle B\] and
\[\angle FCB=\angle FCA\]
that means angle C is bisected by CF.
NOW,
\[\frac{AB}{AC}=\frac{BD}{DC}\]
\[\frac{AB}{BC}=\frac{BD}{DC}=\frac{AE}{CE}\]
\angle B is bisected by BE.
in triangle AFC and AEB,
\angle A is common.\angle AFC and \angle BEA are equal.
so,[\angle ABE=\angle ACF\]
or, \[\frac{1}{2}\angle B=\frac{1}{2}\angle C\]
that means,\[\angle B=\angle C\]
so,\[\angle A=\angle B=\angle C\]
so the ans is 60.

$CF$ isn't Perpendicular bisector
Alternative solution:
Let the intersection point of them be $H$, Perpendicular bisector of $AB$ be $HF$ and altitude from $B$ to $AC$ be $BD$.

Now in $BF=AF$ , $HF$ common side , $\angle{BFH}=\angle{AFH}=90$.

So, $\triangle{AFH}=\triangle{BFH}$ [RHS].

So $\angle{FBH}=\angle{FAH}=\angle{DAH}$. So now in $\triangle{ABD}$

$=>\angle{ABD}+\angle{ADB}=\angle{DAB}=180$

$=>\angle{FBH}+\angle{FAH}+\angle{DAH}+90=180$

$=>3\angle{FAH}=90$

$=>\angle{FAH}=30$

Now $\angle{BAC}=2.\angle{FAH}=2.30=\fbox{60}$