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#1 h Aug 15, 2016, 6:30 PM • 1 Y

Y by Adventure10

PP (NMO Belgium 2004). Let $\triangle ABC$ with $c>b\ ,$ $D\in (AC$ so that $BD=CD$ and

$E\in (BC)\ ,$ $F\in AB$ so that $EF\parallel BD$ and $G\in AE\cap BD\ .$ Prove that $\widehat{BCF}\equiv\widehat{BCG}\ .$

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#2 h Aug 15, 2016, 8:53 PM • 2 Y

Y by Adventure10, Mango247

$D \in [AC]$ extended?

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#3 h Aug 16, 2016, 9:07 AM • 1 Y

Y by Adventure10

any solution please

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Arab

#4 h Aug 16, 2016, 5:20 PM • 2 Y

Y by Adventure10, Mango247

Let $H$ be a point on $AD$ such that $GH\parallel BC$ .

Now $FE\parallel BD\implies \frac{EF}{GB}=\frac{AE}{AG}=\frac{EC}{GH}\implies \triangle CEF\sim \triangle HGB$ .

$BCHG$ is an isosceles trapezoid since $BD=CD$ . Hence we have that $\angle BCF=\angle BHG=\angle BCG$ .

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#5 h Aug 16, 2016, 6:07 PM • 1 Y

Y by Adventure10

Arab's solution is very nice. Below is a sketch of a not so clever solution:

Let $CF$ meet $GD$ at $F'$ . By some angle chasing we observe that we are done if $DC$ is tangent to the circumcircle of $\triangle GCF'$ , which is equivalent to $DB^2=DF'\cdot DG$ . Finding the lengths $DF',DB$ and $DG$ in terms of $AB,BC,CA$ is not too hard, and then we just verify that we do in fact have equality above.

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#6 h Aug 17, 2016, 2:08 AM • 2 Y

Y by Adventure10, Mango247

Thank you for your proofs. See here my proof which is more intricate (with harmonic division).

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#7 h Aug 17, 2016, 10:36 AM • 2 Y

Y by Adventure10, Mango247

Arab wrote:

Let $H$ be a point on $AD$ such that $GH\parallel BC$ .

Now $FE\parallel BD\implies \frac{EF}{GB}=\frac{AE}{AG}=\frac{EC}{GH}\implies \triangle CEF\sim \triangle HGB$ .

$BCHG$ is an isosceles trapezoid since $BD=CD$ . Hence we have that $\angle BCF=\angle BHG=\angle BCG$ .

the proof is very nice ,thanks

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