Triangle related question

My question is:

In $\Delta ABC$, let $AE$ be the angle bisector of $\angle A$. If $\frac{1}{AE} = \frac{1}{AC} + \frac{1}{AB}$ then prove that $\angle A = 120^\circ$.

What I tried: I extended side $AB$ and took a point $M$ on it such that $AC$ is congruent to $AM$. Then I proved that $AE$ is parallel to $MC$. I was trying to prove that $\triangle AMC$ is equilateral so that I get $\angle MAC=60^\circ$. But I am not able to prove it.

Any help to solve this question would be greatly appreciated!

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Draw out the figure... think about the sine rule. – user33056 Jun 5 '12 at 23:32

Mark point $D$ on side $AC$ such that $AE=AD$. Rewrite the given relation as follows: $$\frac{1}{AE}-\frac{1}{AC}=\frac{1}{AB}$$ $$\frac{AC-AE}{AE\cdot AC}=\frac{1}{AB}$$ $$\frac{AC-AD}{AD\cdot AC}=\frac{1}{AB}$$ $$\frac{DC}{AD}=\frac{AC}{AB}$$ Now by the angle bisector theorem: $$\frac{BE}{EC}=\frac{AB}{AC}$$ Combining we obtain: $$\frac{DC}{AD}=\frac{EC}{BE}$$ Therefore, $$\frac{AC}{DC}=\frac{AD+DC}{DC}=1+\frac{AD}{DC}=1+\frac{BE}{EC}=\frac{BE+EC}{BE}=\frac{BC}{EC}$$ Hence $\Delta ABC$ is isometric to $\Delta CDE$. Hence $\angle CDE=\angle CAB$.

Now $AE=AD$ hence in $\Delta AED$ we have $\angle AED = \angle ADE=\beta$. $\angle DAE = \frac{1}{2}\angle CAB = \frac{1}{2} \angle CDE = \alpha$. We obtain: $$\angle DAE + \angle ADE + \angle AED = \alpha + 2\beta = \pi$$ $$\angle CDE +\angle ADE = 2\alpha + \beta = \pi$$ Solving we find $\alpha = \beta$, hence $\angle DAE = \angle ADE = \angle AED =\frac{\pi}{3}$, hence $\angle CAB = \frac{2\pi}{3}$.

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Excellent answer!!! – mgh Jun 6 '12 at 6:18

The Sine rule is a good way to go.

Further hint: Use the sine rule on the triangles ABE and AEB on the angles b, e1, e2 and c (see my crude drawing) . What can you then say about the sides AB and AC? Are they equal? (use $\frac{1}{AE}=\frac{1}{AC}+\frac{1}{AB}$)

What does that mean about the triangles ABE and AEB? (SAS)

What's the angle e2 then? What kind of triangle is AEC?

This hopefully should do it. Good luck!

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