1. **Problem statement:** Given an acute angle $\widehat{xOy} > 50^\circ$, points $A$ on ray $Ox$ and $B$ on ray $Oy$ such that $OA = OB$. $H$ is the midpoint of segment $AB$. 2. **Part (a): Prove $\triangle OAH = \triangle OBH$.** - Since $H$ is midpoint of $AB$, $AH = HB$. - $OA = OB$ by given. - $OH$ is common side. - By Side-Side-Side (SSS) criterion, $\triangle OAH = \triangle OBH$. 3. **Part (b): On ray $OH$, take point $M$ such that $OM > OH$. Prove $AM = MB$.** - Since $M$ lies on $OH$ extended, and $AB$ is segment with midpoint $H$, line $OM$ is along $OH$. - Triangles $OAH$ and $OBH$ are congruent, so $\angle OAH = \angle OBH$. - Because $M$ is on $OH$ extended, $AM$ and $MB$ are symmetric with respect to $OH$. - Hence, $AM = MB$. 4. **Part (c): Through $M$, draw line parallel to $AB$ intersecting $Ox$ at $E$ and $Oy$ at $K$. Prove $OH \perp EK$ and $OM$ is perpendicular bisector of $EK$.** - Since $ME \parallel AB$ and $MK \parallel AB$, $EK$ is parallel to $AB$. - $OH$ is median and altitude in $\triangle AOB$ due to symmetry. - $OH$ is perpendicular to $EK$ because $EK \parallel AB$ and $OH$ is perpendicular to $AB$. - $OM$ lies on $OH$ and passes through midpoint $H$ of $AB$, so $OM$ bisects $EK$. - Therefore, $OM$ is perpendicular bisector of $EK$. **Final answers:** - (a) $\triangle OAH = \triangle OBH$ by SSS. - (b) $AM = MB$ by symmetry about $OH$. - (c) $OH \perp EK$ and $OM$ is perpendicular bisector of $EK$.