altitudes, equal segments given, // wanted (2003 Zhautykov City MO 8-9 P2)

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altitudes, equal segments given, // wanted (2003 Zhautykov City MO 8-9 P2)

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#1 h Aug 23, 2019, 6:29 PM • 1 Y

Y by Adventure10

Let $AM$ and $BN$ be the altitudes of an acute-angled triangle $ABC$ ( $\angle ACB \ne 45 {} ^\circ$ ). The points $K$ and $T$ are marked on the rays $MA$ and $NB$ , respectively, so that $MK = MB$ and $NT = NA$ . Prove that $KT \parallel MN$ .

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#2 h Nov 27, 2020, 7:15 PM

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Let $H$ be the orthocentre of $ABC$ . Since $ANMB$ is cyclic, we have that $AHN\sim BMH$ , hence
$\begin{align*} \frac{HN}{HM}&= \frac{NA}{MB}\\ &=\frac{NT}{MK}. \end{align*}$

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