Point $X$ is chosen inside the non trapezoid quadrilateral $ABCD$ such that $\angle AXD +\angle BXC=180$. Suppose the angle bisector of $\angle ABX$ meets the $D$-altitude of triangle $ADX$ in $K$, and the angle bisector of $\angle DCX$ meets the $A$-altitude of triangle $ADX$ in $L$.We know $BK \perp CX$ and $CL \perp BX$. If the circumcenter of $ADX$ is on the line $KL$ prove that $KL \perp AD$. Proposed by Alireza Dadgarnia
Problem
Source: Iranian TST 2021, second exam day 2, problem 2
Tags: geometry, trapezoid, angle bisector, circumcircle
ilya_math
03.06.2021 21:55
here is my solution. I dont know latex very much so i write it on a paper. sorry for my bad handwriting
Attachments:
geo p5.pdf (263kb)
Brian_Xu
19.12.2021 19:34
My solution is partially inspired by above. Let $E = CX \cap AB$ and $F = BX \cap CD$. Let $O$ be the circumcenter, $H$ be the orthocenter, and $R$ be the circumradius of triangle $AXD$. Angle chasing gives us $\angle EXF = 180 - \angle AXD$, $\angle BEX = \angle BXE = \angle CXF = \angle XFC$, $\angle AEX = \angle XFD = 180 - \angle AXD$. $CL$ is the perpendicular bisector of $XF$, and $BK$ is the perpendicular bisector of $XE$. Let M be a point on the $A$-altitude of triangle $ADX$ such that $OM \perp AX$ and $N$ be a point on the $D$-altitude of triangle $ADX$ such that $ON \perp DX$. Then $ONHM$ is a parallelogram. It's seay to see that $M$ is the circumcenter of triangle $AEX$ and $N$ is the circumcenter of triangle $DFX$, which implies $B, M, K$ are colinear and $C, N, L$ are colinear. We also know that <$NL$, $MK$> = $\angle AXD$. Now we only consider points $H, M, N, O, L, K$. We are given that $L, O, K$ are colinear, and we need to proof $LK \perp AD$, which $\Leftarrow \frac{HK}{HL} = \frac{sin \angle D}{sin \angle A}$. The rest is trig bash, and we only consider the case $\angle A > \angle X > \angle D$. We have $HM = \frac{R\cdot sin(\angle A-\angle X)}{sin \angle X}$, $HN = \frac{R\cdot sin(\angle X-\angle D)}{sin \angle X}$. We also have $\triangle LMO\sim\triangle ONK$, which gives us $ML\cdot NK = MO\cdot NO$. Let $x = \angle HKM$, $y = \angle HMK$, then $x+y+\angle X=180$, and $\angle LNO=y-\angle X$. Calculate $MK, NK$.
$MK = HN\cdot \frac{sin \angle HNL}{sin \angle HLN}-HM = \frac{R}{sin \angle X} \cdot (sin (\angle X - \angle D)\cdot \frac{sin y}{sin (y - \angle X)} - sin(\angle A - \angle X))$, and $NK = \frac{R}{sin \angle X} \cdot (sin (\angle A - \angle X)\cdot \frac{sin y}{sin (y + \angle X)} - sin(\angle X - \angle D))$.
Plug them into $ML\cdot NK = MO\cdot NO$.
$(\frac{R}{sin \angle X} \cdot (sin (\angle X - \angle D)\cdot \frac{sin y}{sin (y - \angle X)} - sin(\angle A - \angle X)))\cdot (\frac{R}{sin \angle X} \cdot (sin (\angle A - \angle X)\cdot \frac{sin y}{sin (y + \angle X)} - sin(\angle X - \angle D))) = \frac{R\cdot sin(\angle A-\angle X)}{sin \angle X}\cdot \frac{R\cdot sin(\angle X-\angle D)}{sin \angle X}$
so
$(sin (\angle X - \angle D)\cdot \frac{sin y}{sin (y - \angle X)} - sin(\angle A - \angle X))\cdot (sin (\angle A - \angle X)\cdot \frac{sin y}{sin (y + \angle X)} - sin(\angle X - \angle D)) = sin(\angle A-\angle X)\cdot sin(\angle X-\angle D)$
Open the brackets and use $sin y = \frac{sin (y - \angle X)+sin (y + \angle X)}{2\cdot cos \angle X}$,
$\frac{sin (\angle A - \angle X)\cdot sin (\angle X - \angle D)}{2\cdot cos \angle X} \cdot (sin (y + \angle X) + sin (y - \angle X)) = sin^2 (\angle X-\angle D) \cdot sin (y + \angle X) + sin^2 (\angle A - \angle X) \cdot sin (y - \angle X)$
From $\frac{HK}{HL} = \frac{sin \angle D}{sin \angle A}$ we have $\frac{sin (y + \angle X)}{sin (y - \angle X)} = \frac{sin \angle A \cdot sin (\angle A - \angle X)}{sin \angle D \cdot sin (\angle X - \angle D)}$, so we only need to check $\frac{sin (\angle A - \angle X)\cdot sin (\angle X - \angle D)}{2\cdot cos \angle X} \cdot (sin \angle A \cdot sin (\angle A - \angle X) + sin \angle D \cdot sin (\angle X - \angle D)) = sin^2 (\angle X-\angle D) \cdot sin \angle A \cdot sin (\angle A - \angle X) + sin^2 (\angle A - \angle X) \cdot sin \angle D \cdot sin (\angle X - \angle D)$ ,which is $sin \angle A \cdot sin (\angle A - \angle X) + sin \angle D \cdot sin (\angle X - \angle D) = 2 \cdot cos \angle X \cdot (sin (\angle X-\angle D) \cdot sin \angle A + sin (\angle A - \angle X) \cdot sin \angle D)$ Now plug in $\angle X = 180 - \angle A - \angle D$, and then some simple calculations can prove that this equation is correct.