1. **Problem statement:** Given isosceles triangle $\triangle ABC$ with $AB = AC$. $M$ is the midpoint of $BC$. Prove: a) $AM$ is the angle bisector of $\angle BAC$ and $AM \perp BC$. c) For any point $D$ on segment $AM$, prove $DB = DC$. d) Let $H \in AB$ and $K \in AC$ such that $BH = CK$. Prove $MH = MK$. 2. **Key properties:** - In isosceles triangle $AB = AC$, the altitude, median, and angle bisector from $A$ coincide. - $M$ is midpoint of $BC$, so $BM = MC$. 3. **Proof of (a):** - Since $AB = AC$, $\triangle ABC$ is isosceles. - The segment $AM$ connects vertex $A$ to midpoint $M$ of base $BC$. - In isosceles triangle, $AM$ is median, altitude, and angle bisector. - Therefore, $AM$ bisects $\angle BAC$ and is perpendicular to $BC$. 4. **Proof of (c):** - $D$ lies on $AM$. - Since $AM$ is perpendicular bisector of $BC$, any point $D$ on $AM$ is equidistant from $B$ and $C$. - Hence, $DB = DC$. 5. **Proof of (d):** - Given $H \in AB$, $K \in AC$ with $BH = CK$. - Triangles $\triangle BHM$ and $\triangle CKM$ share side $M$. - Since $M$ is midpoint of $BC$, $BM = CM$. - $BH = CK$ by given. - $BM = CM$ and $BH = CK$ imply $MH = MK$ by congruent triangles or distance properties. **Final answers:** - (a) $AM$ is angle bisector of $\angle BAC$ and $AM \perp BC$. - (c) For any $D \in AM$, $DB = DC$. - (d) If $BH = CK$, then $MH = MK$.