Math Olympiad Preparation Geometry Question

Question

Let $ABC$ be a triangle with $\angle{C} = 60^{\circ}$. If the bisectors of $\angle{CAB}$ and $\angle{ABC}$ meet the opposite sides at $P, Q$, prove that $$AB = AQ + BP$$

I am stuck on this question, I have tried a few things like the bisector theorem and such but nothing seems to work. There is an outline of my reasoning below, hope someone can help.

$\frac{BP}{\sin(\alpha)}=\frac{AB}{\sin(60º+\alpha)}$

$BP=\frac{AB \sin(\alpha)}{\sin(60º+\alpha)}$

$\frac{AQ}{\sin(60º-\alpha)}=\frac{AB}{\sin(120º-\alpha)}$

$ AQ=\frac{AB \sin(60º-\alpha)}{\sin(120º-\alpha)}$

$BP + AQ = \frac{AB |\sin(\alpha)}{\sin(60º+\alpha)} + \frac{AB \sin(60º-\alpha)}{\sin(120º-\alpha)}$

Factorising $AB$ and making $\sin(60º+\alpha) = \sin(120º-\alpha) $ $$BP + AQ = AB(\frac{\sin(\alpha) + \sin(60º-\alpha)}{\sin(60º+\alpha)}) $$ So the factor of $AB$ should be equal to 1 for the equation to hold but even when adding the sines in the numerator with the rule for adding sines I still dont see how it is possible for the factor to be equal to 1 for an arbitrary $\alpha$. $$BP + AQ = AB(\frac{\sin(\alpha-30º)} {\sin(60º+\alpha)})$$

Answer 1

$\sin \alpha+\sin(60^\circ-\alpha)=2\sin30^\circ\cos(\alpha-30^\circ)=\cos(\alpha-30^\circ)=\sin(60^\circ+\alpha)$

Answer 2

Hint for an alternative (synthetic) proof:

• Prove $CQIP$ is cyclic and
• that triangles $QQ'B$ and $QPB$ are congruent where $Q'$ is a reflection of $Q$ over $AI$.

Answer 3

if you look for a geometric solution, here is a simple one:

The bisectors AP and BQ cross at a point, let's denote it by O. we have:

$\angle BOA= 90 +\frac {60}{2}=120$

where $60=\angle ACB$.

Now draw the bisector of $\angle ABO$ to cross AB at D. We have:

$\angle DOA=120/2=60=\angle AOQ$

Therefore $AQ=AD$ becase triangles AOQ and AOD are similar(two angles are equal) and one common side AO. Similarly we have:

$\angle DOB=120/2=60=\angle BOP$

There for $BD=BP$.

Hence :

$AB=AD +BD=AQ+BP$