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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
up vote 1 down vote accepted

Mark point $D$ on side $AC$ such that $AE=AD$. enter image description here 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!

enter image description here

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