$AP\perp BC$

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Phlembac Adib Hasan
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$AP\perp BC$

Unread post by Phlembac Adib Hasan » Thu Apr 25, 2013 9:51 pm

Let $ABC$ be a triangle with incenter $I$. Points $M$ and $N$ are the midpoints of side $AB$ and $AC$, respectively. Points $D$ and $E$ lie on lines $AB$ and $AC$, respectively, such that $BD = CE = BC$. Line $\ell _1$ pass through $D$ and is perpendicular to line $IM$. Line $\ell _2$ passes through $E$ and is perpendicular to line $IN$. Let $P$ be the intersection of lines $\ell _1$ and $\ell _2$. Prove that $AP\perp BC$.
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zadid xcalibured
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Re: $AP\perp BC$

Unread post by zadid xcalibured » Sat Apr 27, 2013 9:18 pm

This problem is just a mere application of only one lemma.The perpendicular lemma.Use the lemma two times only.
:mrgreen:

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zadid xcalibured
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Re: $AP\perp BC$

Unread post by zadid xcalibured » Sat Apr 27, 2013 11:32 pm

Let the touchpoint of the incricle with $BC$ be $F$.
As $DP \perp IM$, $IP^2-PM^2=ID^2-DM^2$
And as $EP \perp IN$, $PN^2-PI^2=EN^2-EI^2$
Summing these two, we get $PN^2-PM^2=(EN^2-DM^2)+(ID^2-IE^2)$
$PN^2-PM^2=(EN^2-DM^2)+(IC^2-IB^2)$ [Note that $ID=IC$,$IE=IC$]
$PN^2-PM^2=(EN^2-DM^2)+(CF^2-BF^2)$ [As $IF \perp BC$]
$PN^2-PM^2=(a-\frac{b}{2})^2-(a-\frac{c}{2})^2+(s-c)^2-(s-b)^2$
$PN^2-PM^2=(\frac{b}{2})^2-(\frac{c}{2})^2$
$PN^2-PM^2=AN^2-AM^2$
$\Rightarrow AP \perp (MN||BC)$

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