Parallelity and equal angles given, wanted an angle equality

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Parallelity and equal angles given, wanted an angle equality

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#1 h Mar 15, 2022, 8:30 PM • 4 Y

Y by teomihai, HWenslawski, son7, oralayhan

Given a convex quadrilateral $ABCD$ such that $m(\widehat{ABC})=m(\widehat{BCD})$ . The lines $AD$ and $BC$ intersect at a point $P$ and the line passing through $P$ which is parallel to $AB$ , intersects $BD$ at $T$ . Prove that
$m(\widehat{ACB})=m(\widehat{PCT})$

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#2 h Mar 15, 2022, 8:33 PM • 3 Y

Y by teomihai, HWenslawski, oralayhan

Let the line passing through $D$ and parallel to $AB$ intersects $BP$ and $AT$ at $K$ and $L$ , respectively. We have
$\frac{|DK|}{|AB|}=\frac{|PD|}{|PA|}=\frac{|PD|}{|PA|}\cdot\frac{|PT|}{|DL|}\cdot\frac{|DL|}{|PT|}=\frac{|PD|}{|PA|}\cdot\frac{|AP|}{|AD|}\cdot\frac{|DL|}{|PT|}=$ $\frac{|PD|}{|AD|}\cdot\frac{|DL|}{|PT|}=\frac{|PT|}{|AB|}\cdot\frac{|DL|}{|PT|}=\frac{|DL|}{|AB|}\Rightarrow |DK|=|DL|$ Also, $\angle DCK=\angle ABC=\angle DKC\Rightarrow |DK|=|DC|$ .
Therefore, $\angle LCK=90^{\circ}$ .
Hence, it suffices to prove that the angle bisector of $\angle ACT$ is $CL$ .
Let $TA\cap PB=M$ . We have
$\frac{|MA|}{|MT|}=\frac{|AB|}{|TP|}=\frac{|AD|}{|DP|}=\frac{|LA|}{|LT|}\Rightarrow (M,L;A,T)=-1$ Also, $\angle LCM=90^{\circ}$ . Hence, we conclude that the angle bisector of $\angle ACT$ is $CL$ .

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#3 h Mar 15, 2022, 9:30 PM • 2 Y

Y by HWenslawski, oralayhan

Let $AB \cap CD = Q$ and $TC \cap AB = E$ . Note that by the given angle condition, $\triangle QBC$ is isosceles with $QB = QC$ . By angle chasing, it suffices to prove $\angle QCA = \angle CEQ$ , or that $QC^2 = QA \cdot QE$ .

By Menelaus and similarity of triangles $\triangle BCE \sim \triangle PCT$ and $\triangle ADB \sim \triangle PDT$ , we have $1 = \dfrac{QA}{QB} \cdot \dfrac{BC}{CP} \cdot \dfrac{PD}{DA} = \dfrac{QA}{QB} \cdot \dfrac{BE}{PT} \cdot \dfrac{PT}{AB} \implies QA \cdot BE = QB \cdot AB.$
Using the fact that $BE = QE - QB$ , this gives $QA \cdot QE = QB^2 = QC^2$ , which implies the result. $\square$

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#4 h Apr 19, 2024, 7:33 AM

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Let $TA\cap BC=G, TA\cap CF=K$ , foot of the altitude from $A$ to $BC$ be $H$ , the perpendicular at $C$ to $BC$ meet $BA$ at $F$ , foot of the altitude from $T$ to $BC$ be $L$ , $BA\cap CD=E, TC\cap AB=R$ , $GD$ meet $BF, TB$ at $M,N$ respectively.
We want to show that $BC$ is the angle bisector of $\angle RCA \iff (R,A;B,F)=-1$
$(R,A;B,F)=(TC,TA;TB,TF)=(C,G;B,TF\cap BC)=(FC,FG;FB,FT)=(K,G;A,T)=(C,G;H,L)$ $\frac{GH}{GL}=\frac{AH}{TL}=\frac{AB}{TP}=\frac{GB}{GP}=\frac{GM}{GN}=\frac{DM}{DN}=\frac{CH}{CL}$ As desired. $\blacksquare$

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#5 h May 8, 2024, 4:25 PM

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$AB\cap CD=Q$
We will use method of moving points.
Take $QBCD$ fixed. Animate $A$ over $QB$ . Denote $R(XY,ZT)$ as reflection of $XY$ to $ZT$ .
$f:A\rightarrow AC\rightarrow R(AC,BC)\rightarrow R(AC,BC)\cap BQ$ $g: A\rightarrow AD\rightarrow AD\cap BC=T\rightarrow T(QB)_{\infty}\cap BD=P\rightarrow PC\cap QB$ $f,g$ has degree $2$ .
$i)A=B$
$f: B\rightarrow BC\rightarrow BC\rightarrow B$
$g: B\rightarrow BD\rightarrow B\rightarrow B\rightarrow B$
So they are same at $A=B$ .

$ii)A=Q$
Let the parallel line from $C$ to $QB$ intersect $BD$ at $S$ .
$f:Q\rightarrow QC\rightarrow Q'C\rightarrow QB_{\infty}$
$g: Q\rightarrow QD\rightarrow C\rightarrow C(QB)_{\infty}\cap BD=S\rightarrow SC\cap QB=QB_{\infty}$
So they are same at $A=Q$ .

Let the parallel from $D$ to $BC$ intersect $QB$ at $E$ and the parallel from $C$ to $BD$ intersect $QB$ at $F$ .
$iii)A=E$
$f: E\rightarrow EC\rightarrow FC\rightarrow F$
$g: E\rightarrow ED\rightarrow BC_{\infty}\rightarrow (BC)_{\infty}(QB)_{\infty}\cap BD=BD_{\infty}\rightarrow F$
Thus $f,g$ are same at $3$ points as desired. $\blacksquare$

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