You sure the problem is correct?

corrected,thanks.

So easy, LK//ST

Let $ABC$ be not is you are creating a contest and K and L are points not is correct your grammatic mistakes

Ferum_2710 wrote: Let $ABC$ be not is you are creating a contest and K and L are points not is correct your grammatic mistakes Let $ABC$ be an acute angled triangle and $K,L$ are points on $AC,BC$ respectively such that $\angle{AKB}=\angle{ALB}$. $P$ is the intersection of $AL$ and $BK$ and $Q$ is the midpoint of segment $KL$. Let $T,S$ be the intersection $AL,BK$ with $(ABC)$ respectively. Prove that $TK,SL,PQ$ are concurrent

Solution : Let $D=KT\cap SL$, $E=PD\cap KL$ and $F=PD\cap ST$. Note that the angle condition implies $AKLB$ is cylic. Therefore $\angle BKL=\angle BAT=\angle BST$ so $KL\parallel ST$. Finally, $$-1=(P,D;E,F)\overset{L}{=}(T,S;P_{\infty},F)$$So $F$ is the midpoint of $ST$ which means $E$ is the midpoint of $KL$ so $E=Q$ and we're done.

Attachments:

My solution during the contest. $A,B,L,K$ are cyclic. $\angle ALK= \angle ABS=\angle ATS \implies LK\parallel TS$ $PQ \cap TS=N$. $PLQ \approx PTN$ and $PQK \approx PNS$ $|TN|=\frac{|PN|}{|PQ|}.|LQ|=\frac{|PN|}{|PQ|}.|QK|=|NS|$ $PQ$ passes through the midpoint of $TS$. $\frac{|PL|}{|LT|}=\frac{|PK|}{|KS|} \implies \frac{|PL|}{|LT|}.\frac{|KS|}{|PK|}=1$ $\frac{|TN|}{|NS|}=1$ By multipliying this $2$ equalities $\frac{|PL|}{|LT|}.\frac{|TN|}{|NS|}.\frac{|SK|}{|KP|}=1$ Because of Ceva, $TK,SL,PQ$ are concurrent.

bin_sherlo wrote: My solution during the contest. $A,B,L,K$ are cyclic. $\angle ALK= \angle ABS=\angle ATS \implies LK\parallel TS$ $PQ \cap TS=N$. $PLQ \approx PTN$ and $PQK \approx PNS$ $|TN|=\frac{|PN|}{|PQ|}.|LQ|=\frac{|PN|}{|PQ|}.|QK|=|NS|$ $PQ$ passes through the midpoint of $TS$. $\frac{|PL|}{|LT|}=\frac{|PK|}{|PS|} \implies \frac{|PL|}{|LT|}.\frac{|PS|}{|PK|}=1$ $\frac{TN}{NS}=1$ By multipliying this $2$ equalities $\frac{PL}{PT}.\frac{TN}{NS}.\frac{SK}{KP}=1$ Because of Ceva, $TK,SL,PQ$ are concurrent. I think you mean $\frac{PL}{LT}$ instead of $PT$ and also I think you mean $\frac{PL}{PK} = \frac{PS}{KS}$ not $PS$

bin_sherlo wrote: My solution during the contest. $A,B,L,K$ are cyclic. $\angle ALK= \angle ABS=\angle ATS \implies LK\parallel TS$ $PQ \cap TS=N$. $PLQ \approx PTN$ and $PQK \approx PNS$ $|TN|=\frac{|PN|}{|PQ|}.|LQ|=\frac{|PN|}{|PQ|}.|QK|=|NS|$ $PQ$ passes through the midpoint of $TS$. $\frac{|PL|}{|LT|}=\frac{|PK|}{|PS|} \implies \frac{|PL|}{|LT|}.\frac{|PS|}{|PK|}=1$ $\frac{TN}{NS}=1$ By multipliying this $2$ equalities but nice solution $\frac{PL}{PT}.\frac{TN}{NS}.\frac{SK}{KP}=1$ Because of Ceva, $TK,SL,PQ$ are concurrent. But very simple and nice solution

Let $Z$ be the intersection of $SL,KT$ Consider that $\angle PBL=\angle LBT,\angle TAC = \angle CAS$ So$\dfrac{PL}{LT} = \dfrac{BP}{BT} = \dfrac{AP}{AS} = \dfrac{PK}{KS}$ Which means that $LK$//$ST$ Finally, by cava we can know the intersection of$PZ,LK$ is the midpoint of $LK$ Hence$TK,SL,PQ$ are concurrent

Ferum_2710 wrote: bin_sherlo wrote: My solution during the contest. $A,B,L,K$ are cyclic. $\angle ALK= \angle ABS=\angle ATS \implies LK\parallel TS$ $PQ \cap TS=N$. $PLQ \approx PTN$ and $PQK \approx PNS$ $|TN|=\frac{|PN|}{|PQ|}.|LQ|=\frac{|PN|}{|PQ|}.|QK|=|NS|$ $PQ$ passes through the midpoint of $TS$. $\frac{|PL|}{|LT|}=\frac{|PK|}{|PS|} \implies \frac{|PL|}{|LT|}.\frac{|PS|}{|PK|}=1$ $\frac{TN}{NS}=1$ By multipliying this $2$ equalities $\frac{PL}{PT}.\frac{TN}{NS}.\frac{SK}{KP}=1$ Because of Ceva, $TK,SL,PQ$ are concurrent. I think you mean $\frac{PL}{LT}$ instead of $PT$ and also I think you mean $\frac{PL}{PK} = \frac{PS}{KS}$ not $PS$ I changed it, thanks.

Easiest way: Note that $AKLB$ is cyclic and so $\angle STL = \angle STA = \angle SBA = \angle KBA = \angle KLP$, so $KL \parallel ST$. Now the desired result follows from Steiner's theorem (the midpoints of bases of a trapezoid are collinear with the intersections of diagonals and lateral sides).