# geometry question

Might be simple for some of you but pls type solution as I couldnt solve

1) In $$\triangle ABC,\ AD,BE,CF$$ are concurrent cevians and $$AD$$ is altitude.

Then prove that $$AD$$ bisects $$\angle FDE$$

Note by Megh Parikh
5 years ago

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Problem 1 is really simple if u know the concept of harmonic bundles. The official solution is completely unmotivated

- 2 years, 11 months ago

by the way megha did you qualified for INMO?

- 5 years ago

yes and not megha, "Megh"

- 5 years ago

i solved the question no. 2.. please see the solution and tell me whether it is correct or not :

in triangle ABC , FM = 2 and MC = 4 thus FM/MC = 1/2 also BE which is the median passes through M. Thus we conclude that M is the centroid of triangle ABC. thus as AD is perpendicular to BC and CF is angle bisector of triangle ABC, therefore triangle ABC is equilateral. thus in triangle ACF ,by Pythagoras theorem we have AB^2 = 36 + AB^2/4 or AB = 12/3^1/2 THUS PERIMETER OF TRIANGLE ABC = 12 TIMES ROOT OF 3

- 5 years ago

very clever. Better than mine solution.

- 5 years ago

correct!

- 5 years ago

My new problem I created due to my misread of Q2

- 5 years ago

Solution to #1 We drop $$FX \perp BC$$ and $$EY \perp BC$$. We need to show that

$$\angle FDA = \angle ADE$$

$$\iff$$ $$\angle FDX = \angle EDY$$

$$\iff$$ $$\tan \angle FDX = \tan \angle EDY$$

$$\iff$$ $$\boxed{\dfrac{FX}{DX} = \dfrac{ EY}{DY}}$$ ... $$(i)$$ . Thus if we can prove that (i) is true, we can claim that $$AD$$ bisects $$\angle FDE$$

Now, Ceva's theorem in $$\triangle ABC$$ we have

$$\dfrac{BD}{DC} \times \dfrac{CE}{EA} \times \dfrac{AF}{FB} = 1$$.... $$(ii)$$

Now, $$\triangle FBX \sim \triangle ABD$$ and $$\triangle ECY \sim \triangle ACD$$ and so we have after some computation, $$FX = \dfrac{BX}{BD} \times AD$$ and $$EY = \dfrac{CY}{CD} \times AD$$.

Therefore, $$\dfrac{FX}{EY} = \dfrac{BX \times CD}{DB \times CY}$$... $$(iii)$$ Coming back to $$ii)$$ and putting $$\dfrac{CE}{EA} = \dfrac{CY}{YD}$$ and $$\dfrac{AF}{FB} = \dfrac{DX}{BX}$$ we have

$$\dfrac{BD}{DC} \times \dfrac{CE}{EA} \times \dfrac{AF}{FB} = 1$$

$$\implies$$ $$\dfrac{BD}{DC} \times \dfrac{CY}{YD} \times \dfrac{DX}{XB} = 1$$

$$\implies$$ $$\dfrac{DX}{DY} = \dfrac{BX \times DC}{CY \times BD}$$ ....$$(iv)$$. Equating $$(iii)$$ and $$(iv)$$, we have

$$\dfrac{DX}{DY} = \dfrac{FX}{EY}$$ and thus we have $$\boxed{\dfrac{FX}{DX} = \dfrac{ EY}{DY}}$$. Thus we proved $$(i)$$ and hence we are done! :)

- 5 years ago

Well, problem 2 has many solutions i think as i got 2 solutions. The first one requires to use the formula of the length of the angle bisector repeatedly. This will finally fetch you that $$\triangle ABC$$ is equilateral. That is the aim of the problem actually. Prove that $$\triangle ABC$$ is equilateral. If u cannot do just post a comment, Il type out the detailed solution for you.

- 5 years ago

One more hint prove triangle QAP AND PAD similar By using cevians theorem and parallel lines

- 5 years ago

Hint for (1): Draw the line $$\ell$$ parallel to $$BC$$ through $$A$$. Suposse $$DE$$ and $$DF$$ meet $$\ell$$ at $$P$$ and $$Q$$.

- 5 years ago

thanks

- 5 years ago