Inequality with bisectors

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Inequality with bisectors

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Source: St Petersburg Olympiad 2015, Grade 10, P3

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#1 h Oct 17, 2017, 1:35 PM • 1 Y

Y by Adventure10

$ABCD$ - convex quadrilateral. Bisectors of angles $A$ and $D$ intersect in $K$ , Bisectors of angles $B$ and $C$ intersect in $L$ . Prove $2KL \geq |AB-BC+CD-DA|$

This post has been edited 1 time. Last edited by RagvaloD, Jun 22, 2018, 9:19 AM

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#2 h Jul 8, 2020, 3:45 PM

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Consider the case $AB+CD\ge AD+BC$ . The other case can be done similarly.
Let $C'\in DC$ such that $KC'\parallel LC$ and $B'\in AB$ such that $B'C'\parallel BC$ (see the first figure). Then the quadrilateral $AB'C'D$ is circumscribed and $KB'$ bisects the angle $AB'C$ thus $B'K\parallel BL$ . Also let the line through $C'$ parallel to $KL$ intersect $CL, CB$ at $J, M$ respectively and let the line through $B'$ parallel to $KL$ intersect $BL, BC$ at $I, N$ respectively. Then
$2KL \geq |AB-BC+CD-DA|\iff$ $2KL\ge BB’+B’C’+C’C+CB\iff$ $2KL\ge BB’+CC’-NB-CM.\quad (*)$ Now note that we can merge the triangles $C’CM$ and $B’BM$ along the sides $C’M$ and $B’N$ (see the second figure). Moreover the points $I$ and $J$ will be merged together in the incenter of the resulted triangle. Hence if $P$ is the touchpoint of $B’B$ with the incircle then
$(*)\iff KL\ge B’P$ which is obvious as $B’J=KL$ .

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