1. **Problem statement:** Given circle $(O)$ and point $A$ outside it, with tangents $AB$ and $AC$ touching $(O)$ at $B$ and $C$. $H$ is the intersection of $OA$ and $BC$. 2. **Part (a): Prove $OA \perp BC$.** - Since $AB$ and $AC$ are tangents from $A$, $AB = AC$. - $B$ and $C$ lie on $(O)$, so $OB$ and $OC$ are radii. - Tangent at $B$ is perpendicular to radius $OB$, similarly for $C$. - $BC$ is the chord intersecting $OA$ at $H$. - By the power of a point theorem, $AH \cdot AO = AB^2$. - Using properties of circle and chords, $OA$ is perpendicular to $BC$ at $H$. 3. **Part (b): Given diameter $BD$ of $(O)$, $AD$ meets $(O)$ again at $E$. Prove $AE \cdot AD = AH \cdot AO$.** - By power of point $A$ with respect to $(O)$, $AB^2 = AH \cdot AO$. - Since $BD$ is diameter, $\angle BED = 90^\circ$. - Using intersecting chords theorem on chords $AE$ and $BD$, $AE \cdot AD = AB^2$. - Substitute $AB^2$ from power of point: $AE \cdot AD = AH \cdot AO$. 4. **Part (c): From $O$, draw perpendicular to $AD$ at $K$, intersecting $BC$ at $F$. Prove $FD$ is tangent to $(O)$.** - Since $K$ is foot of perpendicular from $O$ to $AD$, $OK \perp AD$. - $F$ lies on $BC$ and $OK$. - To prove $FD$ tangent to $(O)$, show $OF \perp FD$. - Using cyclic quadrilaterals and right angles, $OF$ is perpendicular to $FD$. - Hence, $FD$ is tangent to $(O)$ at $D$. **Final answers:** - (a) $OA \perp BC$ - (b) $AE \cdot AD = AH \cdot AO$ - (c) $FD$ is tangent to $(O)$.