1. **Problem statement:** Given isosceles triangle $\triangle ABC$ with $AB = AC$. $M$ is the midpoint of $BC$. Prove: a) $\triangle ABM$ is the angle bisector of $\angle BAC$ and $AM \perp BC$. c) For any point $D$ on segment $AM$, prove $DB = DC$. d) Points $H \in AB$ and $K \in AC$ satisfy $BH = KC$. Prove $MH = MK$. 2. **Key properties and formulas:** - In isosceles triangle $AB = AC$, the median $AM$ to base $BC$ is also the altitude and angle bisector. - Midpoint $M$ divides $BC$ into two equal segments: $BM = MC$. - Distance formula and congruent triangles criteria (SSS, SAS) will be used. 3. **Proof of (a):** - Since $AB = AC$, $\triangle ABC$ is isosceles. - $M$ is midpoint of $BC$, so $BM = MC$. - By the properties of isosceles triangles, median $AM$ is also the altitude and angle bisector. - Therefore, $AM \perp BC$ and $AM$ bisects $\angle BAC$. 4. **Proof of (c):** - Let $D$ be any point on $AM$. - Consider triangles $DBM$ and $DCM$. - $BM = CM$ (since $M$ is midpoint). - $DM$ is common side. - $\angle BMD = \angle CMD = 90^\circ$ (since $AM \perp BC$). - By RHS (Right angle-Hypotenuse-Side) congruence, $\triangle DBM \cong \triangle DCM$. - Hence, $DB = DC$. 5. **Proof of (d):** - Given $H \in AB$, $K \in AC$ with $BH = KC$. - Consider triangles $MHB$ and $MKC$. - $MB = MC$ (since $M$ is midpoint). - $BH = KC$ (given). - $AM$ is angle bisector, so $\angle HMB = \angle KMC$. - By SAS congruence, $\triangle MHB \cong \triangle MKC$. - Therefore, $MH = MK$. **Final answers:** - (a) $AM$ is the angle bisector of $\angle BAC$ and $AM \perp BC$. - (c) For any $D$ on $AM$, $DB = DC$. - (d) If $BH = KC$, then $MH = MK$.