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 \(\angle FDE\)

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

thanks

Log in to reply

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! :)

Log in to reply

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

Log in to reply

correct!

Log in to reply

very clever. Better than mine solution.

Log in to reply

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

Log in to reply

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.

Log in to reply

My new problem I created due to my misread of Q2

Log in to reply

by the way megha did you qualified for INMO?

Log in to reply

yes and not megha, "Megh"

Log in to reply

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

Log in to reply