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 \(\angle FDE\)

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## Comments

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TopNewesti 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

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very clever. Better than mine solution.

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correct!

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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! :)

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

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thanks

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

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by the way megha did you qualified for INMO?

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yes and not megha, "Megh"

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My new problem I created due to my misread of Q2

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Comment deleted Jan 27, 2014

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

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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.

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Comment deleted Jan 27, 2014

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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 ,

\(\dfrac{\sqrt{3}}{2} \times a = 6\)

\(\implies\) \(3a = \boxed{12 \sqrt{3}}\)

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

Similar case in ADC.

I had already gone by this approach but couldn't solve equations.

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

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