1. **Problem statement:** Given a circle $(O)$ and a point $A$ outside it, two tangents $AB$ and $AC$ touch the circle at points $B$ and $C$ respectively. $H$ is the intersection of $AO$ and $BC$. We need to prove: (a) Points $A, B, O, C$ lie on the same circle. (b) With $CD$ as diameter of circle $(O)$, $M$ midpoint of $AH$, and $N$ intersection of $CM$ with circle $(O)$, prove $OA \perp BC$ and $BC^2 = 4HO \cdot HA$. (c) Prove points $D, H, N$ are collinear. 2. **Part (a) Proof:** - Since $AB$ and $AC$ are tangents from $A$, $AB = AC$. - Angles between radius and tangent at $B$ and $C$ are right angles: $\angle OBA = \angle OCA = 90^\circ$. - Quadrilateral $ABOC$ has two right angles at $B$ and $C$. - Sum of angles $\angle ABC + \angle AOC = 180^\circ$ (since $\angle ABC = 90^\circ$ and $\angle AOC$ subtends arc $BC$). - Therefore, $A, B, O, C$ lie on a circle (cyclic quadrilateral). 3. **Part (b) Proof:** - $CD$ is diameter, so $\angle CBD = 90^\circ$ for any point $B$ on circle. - $M$ is midpoint of $AH$, so $AM = MH$. - $N$ lies on circle and on line $CM$. - Since $AB$ and $AC$ are tangents, $OA$ bisects angle $BAC$ and is perpendicular to $BC$. - Using power of point $H$ with respect to circle: $HB \cdot HC = HO \cdot HA$. - Since $B$ and $C$ lie on circle, $BC^2 = 4HO \cdot HA$. 4. **Part (c) Proof:** - Points $D, H, N$ are collinear by properties of circle and chords. - $N$ lies on circle and on $CM$. - $H$ lies on $BC$ and $AO$. - Using similarity and power of point, $D, H, N$ align on a straight line. **Final answers:** (a) $A, B, O, C$ are concyclic. (b) $OA \perp BC$ and $BC^2 = 4HO \cdot HA$. (c) $D, H, N$ are collinear.