Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My question is: In Triangle ABC , let AE be the angle bisector of angle A. If 1/AE = 1/AC + 1/AB , then prove that angle A = 120 degrees.

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 degrees.But i am not able to prove it.

Any help to solve this question would be greatly appreciated.

share|improve this question
2  
What have you tried? –  Sean Jun 5 '12 at 21:40
    
Draw out the figure... think about the sine rule. –  user33056 Jun 5 '12 at 23:32

2 Answers 2

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 $\angle CAB = \frac{2\pi}{3}$.

share|improve this answer
    
Excellent answer!!! –  meg_1997 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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.