Let $ABC$ be a triangle . The internal bisector of $ABC$ intersects the side $AC$ at $ L$ and the circumcircle of $ABC$ again at $W \neq B.$ Let $K$ be the perpendicular projection of $L$ onto $AW.$ the circumcircle of $BLC$ intersects line $CK$ again at $P \neq C.$ Lines $BP$ and $AW$ meet at point $T.$ Prove that $$AW=WT.$$
Problem
Source: MEMO 2018 T6
Tags: geometry, circumcircle, MEMO 2018
Tintarn
05.09.2018 12:22
First Claim: $ABKP$ is cyclic.
Proof of First ClaimAngle chasing:
\[\angle AKP+\angle PBA=\angle AKC+\frac{\beta}{2}+\angle PBL=\angle AKC+\frac{\beta}{2}+\angle PCL=\angle AKC+\frac{\beta}{2}+\angle KCA=180^\circ+\frac{\beta}{2}-\angle CAK=180^\circ\]where the last equality is since $\angle CAK=\angle CAW=\angle CBW=\angle CBL=\frac{\beta}{2}$.
Motivation for what followsWe want to prove that $W$ is the midpoint of $AT$. Since $AW=AT$ this is equivalent to showing that $C, T, A$ are on a common circle around $W$ i.e. that $\angle TCA=90^\circ$ by Thales. However, we can't really do angle chasing with the angle $\angle TCA$ since $T$ and $C$ are not really related. Rather, $T$ is defined as an intersection of two lines (namely, $AW$ and $BP$) so it is more natural to consider the line through $C$ perpendicular to $AC$ and prove that the three lines coincide in one point (which we then already know to be $T$).
However, for proving that three lines have a common point one common trick is to find three circles such that these lines are the radical axes of two of these circles each. Fortunately, we already have two such circles: The circumcircles of $ABPK$ and $BCP$ have the radical axis $BP$.
If we suggestively draw the circle through $A, K, C$ we find that it intersects the lines $KL$ and $CT$ and the circle $(BPC)$ so that is a good one. This motivates the following definition.
Let $D$ be the intersection of the circle $(BPC)$ with the line $KL$.
Second Claim: $AKCD$ is cyclic.
Proof of Second claimSince $\angle AKD=\angle AKL=90^\circ$ it suffices to prove that $\angle ACD=90^\circ$. But this again is just angle chasing. We have
\[\angle ACD=\gamma+\angle BCD=\gamma+\angle BLD=\gamma+270^\circ-\angle LBA-\angle BAK=\gamma+270^\circ-\alpha-\frac{\beta}{2}-\frac{\beta}{2}=90^\circ\]
So the lines $AK, BP$ and $CD$ are radical axes of three circles, hence have a common point which then must be $T$. Hence $C, D, T$ are collinear. But $\angle ACD=\angle AKD=90^\circ$, hence $\langle TCA=90^\circ$. Hence $A, C, T$ are on a Thales Circle. Since $AW=WC$ by definition of $W$, this implies that $W$ is the midpoint of $AT$ as desired.
MarkBcc168
05.09.2018 12:44
Problem : Let $ABC$ be a triangle. The angle bisector of $\angle BAC$ meets $BC$ and $\odot(ABC)$ at $D, M$ respectively. Let $X$ be the projection from $D$ onto $BM$ and $\odot(ADC)$ meets $CX$ at $Y$. Line $AY$ and $BM$ meet at point $T$. Prove that $BM = MT$. Solution : Let $K$ be the midpoint of $BC$. Notice that $DXMK$ is concyclic thus $\angle XKC = 180^{\circ} - \angle DMX = \angle AMT$. Moreover, $\angle MAT= \angle XCK$, we get $\triangle AMT\sim\triangle CKX$. Thus combining with $\triangle BKX\sim\triangle BMD$ yields $$\frac{MT}{MA} = \frac{KX}{KC} = \frac{KX}{MD}\cdot\frac{MD}{KC} = \frac{BK}{BM}\cdot\frac{MD}{KC} = \frac{MD}{MB} = \frac{MB}{MA}$$hence we are done.
lazizbek42
12.01.2022 16:37
Sinus law: $$\frac{sin \angle ABW}{ sin \angle WBT} = \frac{sin \angle BAW}{sin \angle BTW}$$
MrOreoJuice
13.01.2022 16:45
Let $TC$ intersect $(BLC)$ at $X$ then Pascal on $CCXLBP$ gives that $W-T- XL \cap PC = K'$, so $PC$ intersects $WT$ at $K'$ which forces $K' \equiv K$ and in particular $K-L-X$. $$\measuredangle CTA = 90^\circ - \measuredangle KXC = 90^\circ - \measuredangle LBC =90^\circ - \measuredangle TAC$$So $\measuredangle ACT =90^\circ$ which means $W$ is the circumcenter of $\triangle ACT$ and hence the result.
dgreenb801
14.01.2022 05:45
Let the circumcircle of $\triangle BLC$ meet $AB$ at $D$. Then $\angle DBL = \angle CBL \implies DL=CL$. Also $\angle DCL = \angle DBL = \angle ABL = \angle CBW = \angle CAW$, so $DC \parallel AW$. Let the parallel through $B$ to $CD$ meet the circumcircle of $\triangle BCD$ at $E$, then $BECD$ is an isosceles trapezoid. Let $CE$ intersect $AT$ at $F$, then $ADCF$ is an isosceles trapezoid so since $DL=CL$, by symmetry $AK=KF$. Now projecting through $B$ onto the circle $(BCD)$, $(A,T;W,AT_{\infty})=(D,P;L,E)$ Projecting through $C$ on $AT$, $(D,P;L,E)=(AT_{\infty},K;A,F)=(A,F;K,AT_{\infty})=1$ (since $AK=KF$). Thus $(A,T;W,AT_{\infty})=1$, so $AW=WT$.
MatBoy-123
14.01.2022 08:15
We start with a claim - Claim - $WC$ is tangent to $(BLC)$- Proof - Note that $\angle LCW = \angle ACW = \angle ABW = \angle CBW = \angle CBL$ $\blacksquare$ Now suppose $TC$ intersect $(LBC)$ again at $Q$ , then by Pascal Theorem of the hexagon $CQLBPC$ , we have $\overline{L-K-Q}$ , so this implies that from angle chase that $QL$ is diameter of $(BLC)$ , which implies $\angle ACT = 90$ ,which implies that $(LCKT)$ is cyclic. Claim - $WT$ is tangent to $(BLT)$. Proof - Note that $$\angle ATL = \angle KTL = \angle KCL = \angle PCL = \angle PBL = \angle TBL$$$\blacksquare$. Now from Power of Point we have $$ WT^2 = WL.WB = WC^2 = WA^2$$, where at last we uses Fact-5 , so $TW = AW$ and we are done. $\blacksquare$
CROWmatician
14.01.2022 09:57
Let $T$ be the reflection of $A$ over $W,$ and let $P'=AT\cap KC$, we´ll show $BLP'C$ is cyclic. To do this note that from equal sides we know $\angle{ACT}$ is right so $LKTC$ is cyclic, then $$\angle{LCK}=\angle{LTK}=\angle{LTW}=\angle{LBT}$$(last quality follows from noticing that $WT$ is tangent to $(BLT)$, because $WL\cdot WB=WA^2=WT^2$), hence $BLP'C$ is cyclic, as desired.
X.Allaberdiyev
25.08.2023 14:37
Let $T'$ be the point such that $\angle ACT'=90$ and let $BT'\cap KC=P'$. Then we have $T'W^2=AW^2=WL*WB$, which means that $\angle WBT'=\angle LT'W$, and since $(L K T' C)$ is cyclic, we have $\angle LT'W=\angle LCK$, which means that $\angle LCP'=\angle LBP'$, then $(B L P' C)$ is cyclic, then $P=P'$ which completes the solution.