Lemma 1. Let $D$ be such a point that $ABDC$ is a parallelogram, then point $P$ lies on the angle bisector of $\angle{BDC}$. Lemma 2. The bisectors of $\angle{BAC}$ and $\angle{BDC}$ are parallel. Therefore points $M,P,D$ are collinear and hence $MC=DC=AB$.

My solution uses barycentric coordinates: (these kind of solutions are usually messy, but this one came out nicely) Let $BK=CL=t$. Then $P(t: c-t: b-t)$. The equation of $AD$ (the interior bisector) is $0x+cy-bz=0$ and its point at infinity is $(b+c: -b : -c)$. Let $N$ s.t. $CN=BA$ (properly oriented). The equation of $PN$ is $(c-t)(x(b-c)+yb-zc)=0$. The case $c=t$ is trivial, so $c \neq t$ and the equation of $PN$ is $(b-c)x+yb-zc=0$ and its point at infinity is $(b+c: -b: -c)$. Thus, $AD \| PN$ as they have the same point at infinity. Conclusion: $M=N$, so $CM=AB$. Sailor's solution is very nice, but I don't think I'll ever understand geometry in order to come up with synthetic solutions like that. ( )

I have a strange solution. WLOG $AB\geq AC$ Let $M$ be a point on ray $CA$ such that $AB=CM$, $l$ is a line parallel through $M$ to angle bisectore of $\angle BAC$ Denote $N, L$ be the intersections of line $l$ and $AB, BC$ respectively. Let's think a point $X$ on edge $ML$. We'll call $P, Q$ as the intersection of $BX, AC$ and $CX, AB$ respectively. And Let $S(X)= CP$, $R(X) = BQ$. It's obvious that $S(X), R(X)$ is the 2-dimension functions of $X$. cuz $S(X)=R(X)$ for $X=L, N$, we can know that two functions are same. Therefore, edge $ML$ is the trace of $P$. $Q.E.D.$

Use Menalaus twice and you are done. Bomb

Could anyone post the solution with Menelaos ?Thank you

Valentin Vornicu wrote: Let $ ABC$ be a triangle and $ K$ and $ L$ be two points on $ (AB)$, $ (AC)$ such that $ BK = CL$ and let $ P = CK\cap BL$. Let the parallel through $ P$ to the interior angle bisector of $ \angle BAC$ intersect $ AC$ in $ M$. Prove that $ CM = AB$. My solution is based on the method of working backward and using Menalaus Theorem. Let the angle bisector of $ A$ meet $ BC$ at $ Y$, $ CK \cap AY = X$. We take $ M'$ on $ CA$ such that $ AB = CM'$ and let $ M'P \cap BC = Z, BZ \cap CA = L'$. Now $ \angle ZM'C = \frac A2$. So, by sin law we get, $ \frac {\sin \frac A2}{\sin \angle M'ZC} = \frac {CZ}{CM'} = \frac {BY}{AB} \Longrightarrow BY = ZC \iff BZ = YC$ Using Menelaus's Theorem on $ \triangle ABY$ and transversal $ KXC$ we get, \[ \frac {AK}{KB} \cdot \frac {BC}{CY} \cdot \frac {YX}{XA} = - 1 \] Using Menelaus's Theorem on $ \triangle CMZ$ and transversal $ LPB$ we get, \[ \frac {CL'}{M'L'} \cdot \frac {M'P}{PZ} \cdot \frac {ZB}{BC} = - 1 \] Also, $ M'Z \parallel AY \Longrightarrow \frac {M'P}{PZ} = \frac {AX}{XY}$ From all these relations we get, \[ \frac {AK}{KB} = \frac {M'L'}{L'C} \iff \frac {AB}{KB} = \frac {M'C}{L'C} \iff KB = L'C \] So, $ L' \equiv L \iff M' \equiv M$ and we are done. $ QED$

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Take $D$ on $(BA$, order $B-A-D$ so that $AD = AC$. Obviously, $PM||CD$. Apply Menelaos in triangle $\triangle ACK$ with transversal $LPB$ and get $\frac {KP}{PC}=\frac{b-k}{c} \ (\ 1\ )$, where $b, c$ and $k$ are the lengths of $AC, AB$ and $BK$ respectively. Next, suppose, w.l.o.g. $AB > AC$. The parallel from $P$ to $CC'$ intersects $AB$ at $M'$ and $\frac{KM'}{M'C'} = \frac{KP}{PC} ( 2 )$. Since $AC' = AC = b$, $KC' = b + c - k$. From (1) and (2) we get $KM' = b - k$, that is, $BM' = b = AC' = AC ( 3 )$, hence $C'X = AB ( 4 )$. Now, extend $PM'$ till it intersects $AC$ at $M$; $MM'CC'$ is an isosceles trapezoid, hence the problem is solved. If $AB < AC$ we get the solution in a similar way. Best regards, sunken rock

Let $ T$ the intersection point of $ AB$ and $ PM$. Obviously $ AT=AM$. Apply Menelaus theorem in triangle $ ABL$ with transversal $ KPC$. $ \frac{CL}{CA}.\frac{KA}{KB}.\frac{PB}{PL}=1$ and finally $ \frac{KA}{CA}=\frac{PL}{PB}$. Apply again Menelaus theorem in triangle $ ABL$ with transversal $ TMP$. $ \frac{PL}{PB}.\frac{TB}{TA}.\frac{MA}{ML}=1$ and finally $ \frac{PL}{PB}=\frac{ML}{TB}$. So $ \frac{KA}{CA}=\frac{LM}{TB}$ which is equivalent to $ \frac{KA}{CL+LM+MA}=\frac{LM}{TA+AK+KB}$ or $ \frac{KA}{KB+LM+TA}=\frac{LM}{TA+KA+KB}$. After a few calculations we have that $ (KA-LM)(KA+LM+TA+KB)=0$ which means $ KA=LM$.

I have solution with projective way: let's take $PM \cap AB =N ,PM\cap BC=Q$ by projecting AB to PM with centre C cross ratios R(M,N,P,Q)=R(A,N,K,B) then PM to AC with centre B R(MNPQ)=R(MALC) so R(A,N,K,B)=R(MALC) then $\frac{ \frac{MA}{MC}}{\frac{LA}{LC}}=\frac{\frac{AN}{AB}}{\frac{KN}{KB}}$ since we know that MA=AN and BK=LC we find that MC*LA=AB*KN the rest of solution is obvious hence MC-LA=BA-KN

Hi Use this : If we make $D$ such that $ABDC$ is a paralellogram then the locus of $P$ will be the angle bisector of $BDC$.