I "solved" the problem in the following way. Let D a point on BC, K be the (second) common point of AC and c(ABD) and L be the (second) common point of AB and c(ACD). Let J be the common point of c(ALK), c(BLD) and c(CDK). As <LAK=<CDK=<CJK, then J lies on CL. If, by ipothesys, <LAJ=<CAD then, since <ALC=<ADC, it results that <ACD=<AJL=<AKL. This means that LK and BC are parallel. From here is immediate that the midpoint of BC lies on AJ. PS The problem I solved shoul, at first insigth, be equivalent to the problem given. Comment on that are wellcome.

We multiply the following relations: $\frac{MB}{MC}=\left(\frac{AB}{AC}\right)^2$ (Steiner's theorem), $LB\cdot AB=MB\cdot BC,MC\cdot BC=KC\cdot AC\Longrightarrow\frac{LB}{KC}=\frac{AB}{AC}\Longrightarrow LK\parallel BC$. Remark. Generally, $\frac{BD\cdot BL }{CD\cdot CK}=\frac{AB}{AC}\Longleftrightarrow \sigma [BLD]=\sigma [CKD].$

My solution: Since < BAM = < DAC, the line AM is the reflection of the line AD in the angle bisector of the angle CAB. But the line AD is a median of triangle ABC; hence, the line AM is a symmedian of triangle ABC. Since a symmedian from a vertex of a triangle divides the opposite side in the ratio of the squares of the two adjacent sides, we have $\frac{BM}{CM} = \frac{BA^2}{CA^2}$. On the other hand, since the points A, C, L and M lie on one circle, the Intersecting Chords Theorem yields $BM\cdot BC = BL\cdot BA$, and since the points A, B, K and M lie on one circle, the Intersecting Chords Theorem yields $CM\cdot CB = CK\cdot CA$. Hence, $\frac{BA^2}{CA^2} = \frac{BM}{CM} = \frac{BM\cdot BC}{CM\cdot CB} = \frac{BL\cdot BA}{CK\cdot CA}$. Consequently, $\frac{BA}{CA} = \frac{BL}{CK}$. This yields KL || BC after Thales. PS. I was asked about how to prove the following fact which I used: darij grinberg wrote: Since a symmedian from a vertex of a triangle divides the opposite side in the ratio of the squares of the two adjacent sides, we have $\frac{BM}{CM} = \frac{BA^2}{CA^2}$. Now I posted a proof of this fact at http://www.mathlinks.ro/Forum/viewtopic.php?t=58872 (post #1 Theorem 5). To make it short, it is a particular case of the Steiner theorem, which tells that if M and N are two points on the line BC such that the lines AM and AN are isogonal cevians with respect to triangle ABC, then $\frac{BM}{CM}\cdot\frac{BN}{CN}=\frac{BA^2}{CA^2}$. Darij

LAK=LMB. KMC=LAk. So LMK=180-2A. We most prove that LM=MK. LM/AC=BM/BA. And MK/AB=MC/AC if LM=MK sow we must prove that BM/MC=AB/AC*AB/AC. BM/MC=(sin BAM*AB)/(sinMAC*AC). sinBAM/sinMAC=sinDAC/sinBAD=(DC*AB)/(DB*AC). BD=DC. So it proved

Bonus Question: Prove my corollary below: $Corollary:$ Let $E$ be the point of intersection of line segments $BK$ and $CL$. Then points $A$, $E$ and $D$ are collinear.

It just so happens that $BK$ and $CL$ intersect at $E$ such that $A$ and $D$ all happen to line up with $E$. This means that $E$, $A$, and $D$ all lie on the same line, which follows that $A$, $E$, and $D$ are collinear by definition.

Ambroseyang wrote: It just so happens that $BK$ and $CL$ intersect at $E$ such that $A$ and $D$ all happen to line up with $E$. Proof?

Let $P$ be $A$ - Humpty point of $\triangle ABC$ We have: $\dfrac{BM}{CM} = \dfrac{AB^2}{AC^2} = \dfrac{BP^2}{CP^2}$ Then: $PM$ is $P$ - symmedian of $\triangle BCP$ So: $\angle{BPM} = \angle{CPD} = \angle{ACB} = \angle{BKM}$ or $B$, $K$, $P$, $M$ lie on a circle Similarly: $L$, $C$, $M$, $P$ lie on a circle Hence: $\angle{BPL} = \angle{BPM} + \angle{LPM} = \angle{ACB} + 180^o - \angle{ACB} = 180^o$ or $B$, $P$, $L$ are collinear Similarly: $C$, $P$, $K$ are collinear So: $KL$ $\parallel$ $BC$

Let $CL$ meet $AD$ at $S$. Claim: $B,S,K$ are collinear. Proof : $\angle SCD = \angle BAM = \angle CAD$ so $DC^2 = DS.DA = DB^2$ so $\angle DBK = \angle MAK = \angle BAD = \angle DBS \implies B,S,K$ are collinear. From Ceva Theorem we have $\frac{AK}{KC} . \frac{CD}{DB} . \frac{BL}{LA} = 1 \implies \frac{AK}{KC} = \frac{AL}{LB}$ so $KL || BC$.