AB + AF=CD+ DE when <B=C=<E=<F, BC//EF, convex ABCDEF

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AB + AF=CD+ DE when <B=C=<E=<F, BC//EF, convex ABCDEF

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#1 h Sep 2, 2020, 4:30 PM • 1 Y

Y by Mango247

In a convex hexagon $ABCDEF$ the angles at the vertices $B, C, E$ and $F$ are equal, and the lines $BC$ and $EF$ are parallel. Prove that $AB + AF=CD+ DE$ .

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#2 h Sep 2, 2020, 10:56 PM

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The idea is simple, but it takes a while to write out.

Extend $AB$ to meet $CD$ at $H$ and also extend $AF$ to meet $DE$ at $G$ .
It is then easy to see by a simple angle chase that $\triangle HBC \sim \triangle GFE$ , and that both are isosceles triangles (so $HB=HC$ and $GF=GE$ ).
Let $\omega_1$ be the circumcircle of $\triangle HAD$ and $\omega_2$ be the circumcircle of $\triangle GAD$ , then since $\angle AHD = \angle AGD$ and both angles intercept a segment of the same length, $\omega_2$ is the reflection of $\omega_1$ over $AD$ .
Let $M$ be the midpoint of minor arc $\widehat{AD}$ of $\omega_1$ , and $N$ be the midpoint of minor arc $\widehat{AD}$ of $\omega_2$ , so $M$ and $N$ are reflections across $AD$ .
Since $\triangle HBC$ is isosceles, the angle bisector of $\angle AHD$ (which is $HM$ ) is perpendicular to $BC$ . Similarly, $GN$ is perpendicular to $EF$ .
Since $BC \parallel EF$ , then $HM \parallel GN$ .
Let $G'$ , $E'$ , and $F'$ be the reflections of $G$ , $E$ , and $F$ across $AD$ , then $G'$ lies on $\omega_1$ .
Also, $\angle HMN = \angle GNM = \angle G'MN$ , so by symmetry, $G'$ is the reflection of $H$ across $MN$ and thus $\triangle HAD \cong \triangle G'DA$ .
Thus $AH=G'D=GD$ and $DH=AG'=AG$ .
Then $AB+AF=(AH-HB)+(AG-GF)=(AH-GF)+(AG-HB)=(GD-GE)+(DH-HC)=DE+CD$ .

This post has been edited 1 time. Last edited by dgreenb801, Sep 3, 2020, 1:17 AM
Reason: Typo.

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#3 h Jul 3, 2021, 11:24 AM • 1 Y

Y by Mango247

If $CD=AB$ the trivial.
WLOG $CD>AB$ .Let $G$ and $H$ be points on $AF$ and $CD$ such that $DG$ parallel to $EF$ and $AH$ parallel to $BC$ , respectively.
Let $I$ be intersections of line $ED$ and $AH$ .
Observe that $\angle{DHI}=\angle{DIH}$ .
So we have $AB=CH$ and $AF=IE$ . Hence
$AB+AF=CH+IE=CH+HD+DE=CD+DE$

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