In a triangle $ABC$, let $AP$ be the bisector of $\angle BAC$ with $P$ on the side $BC$, and let $BQ$ be the bisector of $\angle ABC$ with $Q$ on the side $CA$. We know that $\angle BAC=60^\circ$ and that $AB + BP = AQ + QB$. What are the possible values ​​of the angles of triangle $ABC$?

(Answer: $\angle ABC = 80^\circ, \angle BCA=40^\circ, \angle BAC=60^\circ$)

My progress:

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The relationships I found:

$AQ+QB = AB+BP\implies b+x=c+a$

$\angle C = 120^\circ -2\theta$

Considering angle bisector $BQ$ of $\triangle ABC$: $\dfrac{c}{b}=\dfrac{AC}{d}$

Similarly considering $AP$: $\dfrac{c}{a}=\dfrac{b+d}{y}$

$\angle 60^\circ +\angle B+\angle C \implies \angle B+\angle C = 120^\circ$

$x = \dfrac{(b+d)\cdot c}{AC+c}$

$x^2 =\dfrac{(AC)c}{bd}$

From sine rule:

$\displaystyle\frac{\sin60}{AC}=\frac{\sin C}{c}=\frac{\sin2\theta}{b+d}=\frac{\sqrt3}{2AC}$

From cosine rule:

$c^2 = AC^2+BC^2-2\cdot AC\cdot BC\cdot \cos\angle C$

$AC^2 = c^2+BC^2-2c\cdot BC\cdot\cos2\theta$



1 Answer 1


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Hints: As can be seen in figure there are two key points you have to show:

1- Q is on perpendicular bisector of BC.

2- Triangle BPI is isosceles.

This is only possible construction.Using bisectors theorem may help.


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