equal angles starting with a parallelogram with perpenducular

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equal angles starting with a parallelogram with perpenducular

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Source: Mexican Mathematical Olympiad 1994 OMM P3

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#1 h Jul 29, 2018, 6:19 AM • 2 Y

Y by Adventure10, Mango247

$ABCD$ is a parallelogram. Take $E$ on the line $AB$ so that $BE = BC$ and $B$ lies between $A$ and $E$ . Let the line through $C$ perpendicular to $BD$ and the line through $E$ perpendicular to $AB$ meet at $F$ . Show that $\angle DAF = \angle BAF$ .

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#2 h Jul 30, 2018, 8:37 PM • 2 Y

Y by Adventure10, Mango247

% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later!
$\begin{array}{l} \left. \begin{array}{l} {\rm{DM}} \bot {\rm{FN}}\\ \\ {\rm{FC}} \bot {\rm{DN}} \end{array} \right\}\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} NC \bot DF\\ \\ {\rm{CT}} \bot CE\mathop {}\limits_{}^{} \Rightarrow {\rm B}\widehat CT = D\widehat CT\\ C\widehat GM = C\widehat MF = {90^{\rm O}}\mathop {}\nolimits_{}^{} \mathop {}\limits_{}^{} \Rightarrow CK \bot AF\\ \\ \mathop {}\limits_{}^{} \left. \begin{array}{l} {\rm T}{\rm{C}} \bot {\rm{CE}}\\ \\ {\rm A}{\rm K} \bot {\rm{C}}{\rm E} \end{array} \right\}\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} CT//AK\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} D\widehat AF = B\widehat AF \end{array}$ % MathType!End!2!1!

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nguyenthanhlong2000

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#3 h Aug 1, 2018, 2:21 AM • 2 Y

Y by Adventure10, Mango247

We should use this lemma :Let triangle ABC. P is a point on segment BC. M, N on AB, AC such that PM//AC, PN//AB. We will have the line through P and perpendicular to MN through a fixed point

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