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Source: Indian RMO 1991 Problem 1

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1592 posts

Rushil

#1 h Oct 15, 2005, 4:13 AM • 3 Y

Y by vankhea, Adventure10, Mango247

Let $P$ be an interior point of a triangle $ABC$ and $AP,BP,CP$ meet the sides $BC,CA,AB$ in $D,E,F$ respectively. Show that $\frac{AP}{PD} = \frac{AF}{FB} + \frac{AE}{EC}.$
Remark

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2105 posts

#2 h Oct 15, 2005, 8:28 PM • 3 Y

Y by vankhea, Adventure10, Mango247

Rushil wrote:

Let $P$ be an interior point of a triangle $ABC$ and $AP,BP,CP$ meet the sides $BC,CA,AB$ in $D,E,F$ respectively. Show that $\frac{AP}{PD} = \frac{AF}{FB} + \frac{AE}{EC}.$

If i am not wrong this is called
Van aubel theorem!..

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1906 posts

#3 h Oct 15, 2005, 8:37 PM • 3 Y

Y by vankhea, Adventure10, Mango247

Indeed! This is Van Aubel theorem. It follows from Menelaus theorem.

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663 posts

deej21

#4 h Oct 15, 2005, 10:11 PM • 3 Y

Y by vankhea, paradox64ce, Adventure10

You can solve it pretty easily, although not very conveniently using mass points. Just make $AF = k\cdot AB$ and $FB = (1-k)\cdot AB$ where $0 < k < 1$ , and similarly for $AC$ , then you can find the weights of all your points and plug them back in and we're done.

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154 posts

#5 h Jan 27, 2012, 2:08 PM • 3 Y

Y by vankhea, Adventure10, Mango247

Another solution

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894 posts

#6 h Oct 7, 2012, 5:03 PM • 3 Y

Y by MihaiT, Adventure10, Mango247

my solution
Through the point $O$ draw the line $(l)//BC$ cuts the sides $AB$ and $AC$ in $M$ and $N$ respectively.
then we have $\frac{AM}{MB}=\frac{AO}{OD}=\frac{AN}{NC}=x$
From my lemma we have
$\frac{BM}{MA}.\frac{AF}{FB}+\frac{CN}{NA}.\frac{AE}{EC}=1$
$\Rightarrow \frac{AF}{FB}+\frac{AE}{EC}=x=\frac{AO}{OD}$

See my Lemma
Click to reveal hidden text

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223 posts

#7 h Oct 17, 2013, 3:20 PM • 2 Y

Y by Adventure10, Mango247

This is a direct application of Ceva's theorem as well.

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aayush-srivastava

137 posts

aayush-srivastava

#8 h Nov 10, 2015, 9:36 AM • 3 Y

Y by Evan-Chen123, Adventure10, Mango247

quick solution is by using areal ratios(i mean areas here not bary)

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problemsolve_AoPS

218 posts

problemsolve_AoPS

#9 h Apr 4, 2016, 7:09 AM • 2 Y

Y by abhijeetkr, Adventure10

Hi, I think the easy solution would be the application of Menelaus Theorem or Ceva's theorem.

Using Menelaus Theorem:

$\frac{AF}{FB} + \frac{AE}{CE} = \frac{AP(CD+BD)}{PD.BC}$ $\implies \frac{AP.BC}{PD.BC}$ $\implies \frac{AP}{PD}$ our desired result.

PS. Sorry, I couldn't write much as I have typed all this from my 5" mobile phone.

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2125 posts

#10 h Nov 4, 2018, 8:24 AM • 2 Y

Y by Adventure10, Mango247

Let $P \equiv (x:y:z) \implies D \equiv (0:y:z) ; E \equiv (x:0:z) ; F \equiv (x:y:0)$
Hence, $\text{RHS } = \frac{AF}{FB}+\frac{AE}{EC} =\frac{y}{x}+\frac{z}{x} = \boxed{\frac{y+z}{x}}$ And, $\text{LHS } =\frac{AP}{PD}=\frac{[APB]}{[PBD]}=\frac{[APB]}{[ABD]-[APB]} \implies \frac{AP}{PD}=\frac{\tfrac{z}{x+y+z}}{\tfrac{z}{y+z}-\tfrac{z}{x+y+z}} =\boxed{\frac{y+z}{x}}$ Which directly implies that LHS=RHS, Hence,
$\frac{AP}{PD} = \frac{AF}{FB} + \frac{AE}{EC}.$

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2502 posts

ATGY

#11 h Apr 12, 2021, 8:10 PM

Y by

We see that:
$\frac{AP}{PD} = \frac{[ABP]}{[BDP]} = \frac{[APC]}{[DPC]} = \frac{[ABPC]}{[BPC]}$ $\frac{AF}{FB} = \frac{[ACF]}{[BCF]} = \frac{[AFP]}{[BFP]} = \frac{[APC]}{[BPC]}$ $\frac{AE}{EC} = \frac{[AEB]}{[CEB]} = \frac{[AEP]}{[CEP]} = \frac{[ABP]}{[BPC]}$ Notice that:
$\frac{[ABP]}{[BPC]} + \frac{[APC]}{[BPC]} = \frac{[ABPC]}{[BPC]}$ This means that $\frac{AP}{PD} = \frac{AF}{FB} + \frac{AE}{EC}$ is true. So, we are done.

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lifeismathematics

1188 posts

lifeismathematics

#12 h Nov 14, 2022, 9:20 AM

Y by

lotus lemma ftw
lotus lemma:-https://artofproblemsolving.com/community/q1h509770p25873315

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Aryamathematics

8 posts

Aryamathematics

#13 h Sep 14, 2023, 2:30 AM

Y by

Can be done by area relation

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894 posts

#14 h Sep 14, 2023, 11:12 AM • 1 Y

Y by MihaiT

See more generalization
https://www.cut-the-knot.org/triangle/FourthIdentityInTriangle.shtml

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