# 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
4 years, 5 months ago

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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

- 4 years, 5 months ago

very clever. Better than mine solution.

- 4 years, 5 months ago

correct!

- 4 years, 5 months 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! :)

- 4 years, 5 months ago

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

- 4 years, 5 months ago

thanks

- 4 years, 5 months ago

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

- 2 years, 3 months ago

by the way megha did you qualified for INMO?

- 4 years, 5 months ago

yes and not megha, "Megh"

- 4 years, 5 months ago

My new problem I created due to my misread of Q2

- 4 years, 5 months ago

Comment deleted Jan 27, 2014

OHHH DEAR! u MISREAD THE PROBLEM ITSELF! :3 iTS NOT $$BE$$ , ITS $$DE$$

- 4 years, 5 months 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.

- 4 years, 5 months ago

Comment deleted Jan 27, 2014

No It will not. Here is the solution. This one is more easier as it just requires basic euclidean geometry. Here's the solution.

We observe that $$N$$ is the midpoint of $$FC$$. And we already have $$E$$ to be the midpoint of $$AC$$. Thus , $$EN \parallel AF$$ and so $$DE \parallel AB$$. Therefore we have $$D$$ to be the midpoint of $$BC$$. Moreover we have $$AD \perp BC$$. Therefore $$\triangle ADB \cong \triangle ADC$$. So we have $$AB = AC$$. Again, $$\triangle AMF \sim \triangle MDN$$ and by equating ratios we have $$\dfrac{AM}{MD} = \dfrac{1}{2}$$ and so M is the centroid of $$\triangle ABC$$ Therefore $$CF$$ is the median as well as the angle bisector. So we have

$$\dfrac{AF}{FB} = \dfrac{AC}{BC}$$ $$\implies$$ $$AC = BC$$ . Therefore, $$AB = BC = AC$$ and therefore $$\triangle ABC$$ is equilateral. Thereofre , if each side = $$a$$ cm , The length of the median is $$\dfrac{\sqrt{3}}{2} \times a$$.So ,

$$\dfrac{\sqrt{3}}{2} \times a = 6$$

$$\implies$$ $$3a = \boxed{12 \sqrt{3}}$$

- 4 years, 5 months ago

I solved the problem in another way. Using the fact that $$CN,CM,CF$$ are the internal angle bisectors of $$\angle C$$ in $$\triangle CED, \triangle CAD, CAB$$ and using the formula that for a triangle $$ABC$$ with usual notations, if AD be the internal angle bisector, we have $$AD^2 = bc[1-\dfrac{a^2}{(b+c)^2}]$$

- 4 years, 5 months ago

This approach maybe right. Do you get same answer?

It is very hard in BED to find NC as BE is also complicated.

- 4 years, 5 months ago

I thought it's quite easy though. U just need to keep calm and the rest is easy! :)

- 4 years, 5 months ago

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

- 4 years, 5 months ago