49. The midpoint of a cevian in an acute triangle.

by Virgil Nicula, Jul 4, 2010, 8:15 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=355743

Enunciation. Let an acute $\triangle ABC$ with orthocenter $H$ , $E\in BH\cap AC$ , $F\in CH\cap AB$ and $D$ for which $\frac {DB}{DC}=\lambda$ . Define $P\in AD\cap EF$

and $T\in (BC)$ for which $TB=DC$ . Prove that $\boxed {PA=PD\Longleftrightarrow (\lambda +1)a^2=b^2+\lambda c^2\Longleftrightarrow (\lambda +1)\cdot TA=a\sqrt {\lambda^2+\lambda +1}}$

Particular cases. If $\lambda =1\implies$ here . If $\lambda =\frac ca\implies$ $\boxed {PA=PD\Longleftrightarrow A=60^{\circ}\Longleftrightarrow(b+c)\cdot AT=2am_a}$ .

Proof. Let $S\in (BC)$ so that $\widehat{SAB}\equiv\widehat {DAC}$ , i.e. rays $[AS$ , $[AD$ are isogonally in $\widehat{BAC}$ . From Steiner's theorem obtain $\frac {SB}{SC}=\frac {c^2}{\lambda b^2}$ . i.e. $\frac {SB}{c^2}=\frac {SC}{\lambda b^2}=\frac {a}{c^2+\lambda b^2}$ . Prove

easily that $\boxed {\ AS=\frac {(\lambda +1)bc}{c^2+\lambda b^2}\cdot AD\ }\ (*)$ . Since $\triangle AEF\sim\triangle ABC$ obtain $\frac {AP}{AS}=\frac {AF}{AC}=\cos A$ , i.e. $AP=AS\cdot\cos A$ . Hence $\boxed {PA=PD}$ $\Longleftrightarrow$ $AD=2\cdot PA$ $\Longleftrightarrow$

$AD=2\cdot AS\cdot\cos A$ $\stackrel{(*)}{\Longleftrightarrow}$ $\frac {c^2+\lambda b^2}{(\lambda +1)bc}=2\cdot \cos A$ $\Longleftrightarrow$ $c^2+\lambda b^2=$ $(\lambda +1)\left(b^2+c^2-a^2\right)$ $\Longleftrightarrow$ $\boxed{(\lambda +1)a^2=b^2+\lambda c^2}\ (1)$ . Apply Stewart's relation to the cevian $AT$ in

$\triangle ABC$ and obtain $(\lambda +1)^2\cdot AT^2+\lambda a^2=(\lambda +1)\left(b^2+\lambda c^2\right)$ . Hence $(1)$ becomes $(\lambda +1)^2a^2=(\lambda +1)^2\cdot AT^2+\lambda a^2$ $\Longleftrightarrow$ $\boxed {(\lambda +1)\cdot TA=a\sqrt {\lambda^2+\lambda +1}}$ .

==============================================================================

Proof of the relation $(*)$ . Denote $\phi =m(\widehat {SAB})=m(\widehat {DAC})$ . Therefore, $\left\|\begin{array}{c} \frac {AS}{\sin B}=\frac {SB}{\sin\phi}\\\\ \frac {AD}{\sin C}=\frac {DC}{\sin \phi}\end{array}\right\|\ \implies$ $\frac {AS}{AD}=\frac bc\cdot\frac {ac^2}{c^2+\lambda b^2}$ $\cdot \frac {\lambda +1}{a}\implies (*)$ .
This post has been edited 20 times. Last edited by Virgil Nicula, Nov 27, 2015, 7:39 AM

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