79. A quadrilateral with many perpendicularities.

by Virgil Nicula, Aug 4, 2010, 7:14 PM

Quote:
In quadrilateral $ABCD$ denote midpoints $M$ , $N$ of $[AB]$ , $]CD]$ respectively and orthocenters $H$ , $K$ of $\triangle AOD$ , $\triangle BOD$ , where $O\in AC\cap BD$ . Prove that $MN\perp HK$ .

Proof.
$\blacktriangleright$ $\left\|\begin{array}{ccc}
DH\perp AC & \Longleftrightarrow & HA^2-HC^2=DA^2-DC^2\\\
AH\perp BD & \Longleftrightarrow & HB^2-HD^2=AB^2-AD^2\end{array}\right\|\ (1)$ . Apply theorem of median in triangles $\left\|\begin{array}{ccc}
\triangle AHB & : & 4\cdot HM^2=2\cdot\left(HA^2+HB^2\right)-AB^2\\\
\triangle CHD & : & 4\cdot HN^2=2\cdot\left(HC^2+HD^2\right)-CD^2\end{array}\right\|$ $\implies$ $4\cdot\left(HM^2-HN^2\right)=2\cdot\left[\left(HA^2-HC^2\right)+\left(HB^2-HD^2\right)\right]-AB^2+CD^2\stackrel{(1)}{=}$ $2\cdot\left(DA^2-DC^2+AB^2-AD^2\right)-AB^2+CD^2$ $\implies$ $\boxed {4\cdot\left(HM^2-HN^2\right)=AB^2-CD^2}\ (3)$ .
$\blacktriangleright$ $\left\|\begin{array}{ccc}
BK\perp AC & \Longleftrightarrow & KA^2-KC^2=BA^2-BC^2\\\
CK\perp BD & \Longleftrightarrow & KB^2-KD^2=CB^2-CD^2\end{array}\right\|\ (2)$ . Apply theorem of median in triangles $\left\|\begin{array}{ccc}
\triangle AKB & : & 4\cdot KM^2=2\cdot\left(KA^2+KB^2\right)-AB^2\\\ 
\triangle CKD & : & 4\cdot KN^2=2\cdot\left(KC^2+KD^2\right)-CD^2\end{array}\right\|$ $\implies$ $4\cdot\left(KM^2-KN^2\right)=2\cdot\left[\left(KA^2-KC^2\right)+\left(KB^2-KD^2\right)\right]-AB^2+CD^2\stackrel{(2)}{=}$ $2\cdot\left(BA^2-BC^2+CB^2-CD^2\right)-AB^2+CD^2$ $\implies$ $\boxed {4\cdot\left(KM^2-KN^2\right)=AB^2-CD^2}\ (4)$ .
$\blacktriangleright$ From $(3)$ , $(4)$ obtain $HM^2-HN^2=KM^2-KN^2 \ \Longleftrightarrow\ HK\perp MN$ .
This post has been edited 3 times. Last edited by Virgil Nicula, Nov 23, 2015, 3:24 PM

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