Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

1) $\triangle ABC: D \in AB; E\in BC$ such that $BD=3AD$, $BE = 4EC$. F is the intersection of AE and CD. Prove that FD = FC (I think we should prove $S_{ACE} = S_{ADE} = \frac{S_{ABE}}{4}$)

2)$\triangle ABC: \widehat{A}=90^o, AM \perp BD$ (AM and BD are medians) . $AB = \sqrt{2}$. Find $S_{ABC}$

3)$\triangle ABC: \widehat{A}=30^0; \widehat{B}=50^0$ Prove $ab = c^2 - b^2$ ($\ AB=c; AC=b; BC=a$)

I think we 'll prove $$\frac{c+b}{c} = \frac{c}{b} => b(b+c)=c^2$$

Please help me, I have more than 10 exercises and have to finish it in one day :pokerface:. Thanks

share|cite|improve this question

1) Make an affine transformation (this does not affect any of the length ratios) such that $AE$ becomes the bisector of $\hat A$. Since the bisector splits the opposite side in the proportion of the adjacent sides, this will be the case iff $AB=4AC$, i.e. $AD=AC$. Again, because the bisector splits the opposite side in the proportion of the adjacent sides, we conclude $DF=FC$.

share|cite|improve this answer
I proved that now It is easy to prove $$S_{ACE} = S_{ADE} = \frac{S_{ABE}}{4}$$, after that, everything is easy too But I haven't done Ex. 2 and 3 yet – Xeing Nov 15 '12 at 4:08
I finished all of the things now :D, Can't believe it was so easy – Xeing Nov 15 '12 at 5:32
Ah, great to hear. Indeed, going via the areas is equivalent to though probably less obscure than making the affine transformation argument (after all the theorem about bisectors I used is readily proved via area comparison). – Hagen von Eitzen Nov 15 '12 at 7:10

Your Answer


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.