Let $M$ be the midpoint of $BC$; it suffices to show that $DM$ is a symmedian in triangle $ACD$. Observe that $\angle BCD=\angle BAC$ so triangles $DCA\sim DBC$. Now, by the Ratio Lemma in triangle $DBC$, $\frac{\sin \angle MDB}{\sin \angle MDC}=\frac{CD}{DB}=\frac{AD}{CD}$. Call $N$ the intersection of $DM$ and $AC$. Then by the Ratio Lemma in triangle $DCA$, $$\frac{NA}{NC}=\left (\frac{AD}{CD}\right) \left (\frac{\sin \angle ADN}{\sin \angle NDC}\right )=\left (\frac{AD}{CD}\right)^2,$$implying the desired result.

Here's my solution: Suppose $M$ is the midpoint of $BC$. Let $DM \cap AC=X$, and let $Y$ be a point on $AC$ such that $DY \parallel BC$. Note that it suffices to show that $DM$ (i.e. $DX$) is the $D$-symmedian of $\triangle DAC$. Taking $\infty_{BC}$ to be the point at infinity on $BC$, we get that $$-1=(B,C;M,\infty_{BC}) \overset{D}{=} (A,C;X,Y) \Rightarrow \frac{CX}{XA}=\frac{CY}{YA}=\frac{BD}{DA}$$where the last statement follows from the fact that $DY \parallel BC$. Then, using $DC^2=DB \cdot DA$, and by Steiner's Ratio Theorem, we have $$\frac{CX}{AX}=\left( \frac{CD}{DA} \right)^2 \Rightarrow DX \text{ is the D-symmedian of } \triangle DAC \text{. } \blacksquare$$

Draw the line pass through $K$ and parallel to $BC$ cuts $AB,AC$ at $R,S$, respectively. By chansing angle it is easy to show that $KR=KS=KA=KC$. finally, by Thales we have $DK$ bisects $BC$. Q.E.D

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.83846291115687, xmax = 27.57141564908325, ymin = -10.665376937613544, ymax = 9.509277084289968; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((6.634419620678282,3.4677875626527555)--(-5.3,-5.02)--(9.283767341548215,-4.964544968288097)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((6.634419620678282,3.4677875626527555)--(-5.3,-5.02), linewidth(2) + rvwvcq); draw((-5.3,-5.02)--(9.283767341548215,-4.964544968288097), linewidth(2) + rvwvcq); draw((9.283767341548215,-4.964544968288097)--(6.634419620678282,3.4677875626527555), linewidth(2) + rvwvcq); draw(circle((1.9828860243655388,-2.626038478079622), 7.666255970963545), linewidth(0.8) + rvwvcq); draw(circle((13.591071520717595,1.0211279109769398), 7.3743575387594955), linewidth(0.4)); draw((xmin, 0.0038025175808938188*xmin + 8.343858474576866)--(xmax, 0.0038025175808938188*xmax + 8.343858474576866), linewidth(0.4) + linetype("4 4") + dtsfsf); /* line */ draw((6.634419620678282,3.4677875626527555)--(13.563030599251736,8.39543213688072), linewidth(0.4)); draw((13.563030599251736,8.39543213688072)--(9.283767341548215,-4.964544968288097), linewidth(0.4)); draw((9.283767341548215,-4.964544968288097)--(4.802820879713037,-1.7400454323116576), linewidth(0.4)); draw((4.802820879713037,-1.7400454323116576)--(6.634419620678282,3.4677875626527555), linewidth(0.4)); draw((xmin, 1.156990288324524*xmin-7.296862546701889)--(xmax, 1.156990288324524*xmax-7.296862546701889), linewidth(0.4) + linetype("4 4")); /* line */ /* dots and labels */ dot((6.634419620678282,3.4677875626527555),dotstyle); label("$A$", (5.785176838447768,4.734211043602746), NE * labelscalefactor); dot((-5.3,-5.02),dotstyle); label("$B$", (-6.510618216321846,-6.069375873452094), NE * labelscalefactor); dot((9.283767341548215,-4.964544968288097),dotstyle); label("$C$", (9.54554134548896,-6.576726640275111), NE * labelscalefactor); dot((13.563030599251736,8.39543213688072),dotstyle); label("$D$", (14.410139874439075,6.972523250174881), NE * labelscalefactor); dot((4.802820879713037,-1.7400454323116576),dotstyle); label("$K$", (3.367799655349858,-1.6225956230621186), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Obviously the tangent at $D$ is parallel to $BC$, hence, $$-1=(A,C;KD \cap \odot (ACD),D) \overset{D}{=} (B,C;KD \cap BC, \infty_{BC})$$

nguyendangkhoa17112003 wrote: Draw the line pass through $K$ and parallel to $BC$ cuts $AB,AC$ at $R,S$, respectively. By chansing angle it is easy to show that $KR=KS=KA=KC$. finally, by Thales we have $DK$ bisects $BC$. Q.E.D You mean of course "line pass through $K$ and parallel to $BC$ cuts $AB, DC$"