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In $\bigtriangleup ABC$, the perpendicular bisector of $AB$ and $AC$ meet at $O$. $AO$ meets $BC$ at $D$. Now, $OD$ = $BD$ = $\dfrac {1}{3}BC$. Find the angles of $\bigtriangleup ABC$.

Are there anyone who solve this in the exam time? Did you @dshasan?

We know that O is the circumcenter. So, $OA=OB=OC$. And $DB=DO$. So, $ \angle DBO=\angle DOB=a$.

Take D' such that $D'C= \frac {1}{3} BC$. So, $DO=DB=DD'=D'C$.

$OB=OC$. So, $\angle OBC= \angle OCB=a$. So by $SAS$ congruence $\triangle OBD \cong \triangle OCD'$.
So, $OD=OD'$, And, $\triangle ODD'$ is equilateral. And $\angle ODD'=2a$. So, $a=30^o$. The rest are calculation.

So, $\angle BOA = 180^o-a=150^o$. So, $\angle OAB=\angle OBA= 15^o$.

And now, $\angle COA = 180^o-3a=90^o$. So, $\angle OAC=\angle OCA= 45^o$.

So, $\angle ABC=30^o+15^o= 45^o$. $\angle BCA= 30^o+45^o=75^o$.

And, $\angle CAB= 45^o+15^o= 60^o$.

Answer $(\angle A,\angle B,\angle C)=(60^o,45^o,75^o)$

@Thamim, yes, I solved this during exam hour.

I am actually kicking me for not trying 8,9 in the exam. They both are easy.

Don't kick yourself. You have 1 year more to be the COC or COO.

How does one actually kick himself?

You are right,Thamim...I also feel like kicking myself for not paying much attention to this problem as I was working on other problems...or else I could have attained a bit more than being 2nd runners up......Now,This problem does need only a bit of algebraic calculations..We may let angle BAO to be y,then we may ensure that both AOB And AOC are isosceles triangles.Then by angle chasing,find angle BOD=angle OBD=angle OCD.Then by the rest angle calculations,derive an equation which follows 3y+5y+4y=180,find y =15..That fulfills our requirements of 60-75-45 angled triangle...

ahmedittihad wrote:How does one actually kick himself?

This is a valid question.