Equivalence of identities in a triangle

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Equivalence of identities in a triangle

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Source: Turkey JBMO TST 2014 P7

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#1 h Jun 21, 2014, 2:14 PM • 2 Y

Y by Amir Hossein, Adventure10

Let a line $\ell$ intersect the line $AB$ at $F$ , the sides $AC$ and $BC$ of a triangle $ABC$ at $D$ and $E$ , respectively and the internal bisector of the angle $BAC$ at $P$ . Suppose that $F$ is at the opposite side of $A$ with respect to the line $BC$ , $CD = CE$ and $P$ is in the interior the triangle $ABC$ . Prove that
$FB \cdot FA+CP^2 = CF^2 \iff AD \cdot BE = PD^2.$

This post has been edited 2 times. Last edited by Amir Hossein, Jun 22, 2018, 4:42 AM
Reason: Removed the bad character.

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mathuz

#2 h Jun 22, 2014, 6:48 AM • 1 Y

Y by Adventure10

are you sure?
$AD\cdot BE=DP^2$

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#3 h Jun 22, 2014, 1:24 PM • 4 Y

Y by mathuz, Amir Hossein, Adventure10, Mango247

Let $E_1$ be on $AC$ such that $BE_1||ED$ . Then $AD*BE=AD*DE_1=PD^2$ hence $\angle{E_1BA}=\angle{E_1PA}=\frac{\angle{B}-\angle{A}}{2}$ hence $AE_1PB$ is cyclic ,then $\angle{PBE_1}=\angle{PAE_1}=\frac{\angle{A}}{2}$ hence $\angle{PBC}=\frac{\angle{B}}{2}$ which means $I=P$ so it is easy to see $FC^2-IF^2=IF^2=FB*FC$ . SO WE ARE DONE.

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#4 h Jun 22, 2014, 4:53 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

By applying Menelaus' Theorem, point $D,E,F$ are collinear then $\frac{DC}{DA} \cdot \frac{FA}{FB} \cdot \frac{EB}{EC}=1.$
We also have $CE=DE$ so $\frac{BE}{BF}= \frac{DA}{FA}$ . Since $AP$ is the bisector of angle $BAC$ so $\frac{BE \cdot DA}{BF \cdot FA}= \left( \frac{DA}{FA} \right)^2= \frac{PD^2}{PF^2}.$
It follows that $PD^2=BE \cdot AD \iff FP^2=FA \cdot FB$ .
Let $I$ be the midpoint of $DE$ then $\angle ICE= \angle ICD$ and $\angle CIE=90^{\circ}$ . Therefore $I$ is the incenter of triangle $ABC$ . Hence $\angle CIE+ \angle BIE= 90^{\circ}+ \angle BIE=\angle BIC= 90^{\circ}+ \frac{\angle A}{2}.$
We obtain $\angle BIE= \frac{ \angle A}{2}= \angle IAB$ . It follows that $\triangle FIA \sim \triangle FBI \; ( \text{A.A})$ . Hence $FI^2=FB \cdot FA$ .
Thus, $PD^2=BE \cdot AD \Leftrightarrow PF^2=FA \cdot FB \Leftrightarrow PF=FI \\ \Leftrightarrow P \equiv I \Leftrightarrow \angle CPF=90^{\circ} \Leftrightarrow FP^2+PC^2=FC^2 \\ \Leftrightarrow FA \cdot FB+CP^2=CF^2.$

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mathuz

#5 h Jun 23, 2014, 12:12 PM • 1 Y

Y by Adventure10

mathuz wrote:

are you sure?
$AD\cdot BE=DP^2$

Let $P$ be any point on the segments $ED$ . Then

$FB\cdot FA+CP^2 = CF^2\:\Longleftrightarrow\: AD\cdot BE = PD\cdot PE .$

The our problem will be particular case.

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#6 h Jul 6, 2014, 11:52 AM • 2 Y

Y by Adventure10, Mango247

shinichiman wrote:

By applying Menelaus' Theorem, point $D,E,F$ are collinear then $\frac{DC}{DA} \cdot \frac{FA}{FB} \cdot \frac{EB}{EC}=1.$
We also have $CE=DE$ so $\frac{BE}{BF}= \frac{DA}{FA}$ . Since $AP$ is the bisector of angle $BAC$ so $\frac{BE \cdot DA}{BF \cdot FA}= \left( \frac{DA}{FA} \right)^2= \frac{PD^2}{PF^2}.$
It follows that $PD^2=BE \cdot AD \iff FP^2=FA \cdot FB$ .
Let $I$ be the midpoint of $DE$ then $\angle ICE= \angle ICD$ and $\angle CIE=90^{\circ}$ . Therefore $I$ is the incenter of triangle $ABC$ . Hence $\angle CIE+ \angle BIE= 90^{\circ}+ \angle BIE=\angle BIC= 90^{\circ}+ \frac{\angle A}{2}.$
We obtain $\angle BIE= \frac{ \angle A}{2}= \angle IAB$ . It follows that $\triangle FIA \sim \triangle FBI \; ( \text{A.A})$ . Hence $FI^2=FB \cdot FA$ .
Thus, $PD^2=BE \cdot AD \Leftrightarrow PF^2=FA \cdot FB \Leftrightarrow PF=FI \\ \Leftrightarrow P \equiv I \Leftrightarrow \angle CPF=90^{\circ} \Leftrightarrow FP^2+PC^2=FC^2 \\ \Leftrightarrow FA \cdot FB+CP^2=CF^2.$

You say $I$ is incenter of triangle $ABC$ . But if you know that we can immediately conclude that $AI$ is angle bisector which means $P=I$ . Your reasons are not enough to conclude that $I$ is incenter.

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Arslan

#7 h Nov 9, 2014, 7:23 AM • 2 Y

Y by Adventure10, Mango247

Can someone clarify why P is the incenter of the triangle ABC?

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