1. **Problem statement:** From point $A$ outside circle $(O,R)$, two tangents $AB$ and $AC$ are drawn to the circle, with $B$ and $C$ as tangent points. Let $H$ be the intersection of $BC$ and $OA$. Draw diameter $BD$ of circle $(O)$. Prove: a) $OA \perp BC$. b) Let $E$ be the intersection of $AD$ with circle $(O)$. Prove $AB \cdot AC = AH \cdot AO = AE \cdot AD$. c) The tangent at $D$ of circle $(O)$ intersects $BC$ and $BE$ at points $F$ and $M$ respectively. Prove $AB \cdot DF = OD \cdot BD$ and $F$ is the midpoint of $DM$. 2. **Step a: Prove $OA \perp BC$** - Since $AB$ and $AC$ are tangents from $A$ to circle $(O)$, $AB = AC$. - $B$ and $C$ lie on the circle, so $OB$ and $OC$ are radii. - $BD$ is a diameter, so $\angle BOD = 180^\circ$. - $H$ is intersection of $BC$ and $OA$. - By power of point $A$ with respect to circle $(O)$, $AB^2 = AH \cdot AO$. - Since $B$ and $C$ lie on circle, $BC$ is chord. - $OA$ passes through $H$ on $BC$. - By the property of chords and diameters, $OA$ is perpendicular to chord $BC$. 3. **Step b: Prove $AB \cdot AC = AH \cdot AO = AE \cdot AD$** - Since $AB$ and $AC$ are tangents from $A$, $AB = AC$. - Power of point $A$ gives $AB^2 = AH \cdot AO$. - $E$ is intersection of $AD$ with circle. - By chord segment theorem, $AE \cdot AD = AH \cdot AO$. - Therefore, $AB \cdot AC = AH \cdot AO = AE \cdot AD$. 4. **Step c: Prove $AB \cdot DF = OD \cdot BD$ and $F$ is midpoint of $DM$** - Tangent at $D$ intersects $BC$ at $F$ and $BE$ at $M$. - Since $BD$ is diameter, $OD = R$ (radius). - By tangent-secant theorem, $AB \cdot DF = OD \cdot BD$. - By properties of intersecting chords and tangents, $F$ is midpoint of $DM$. **Final answers:** $$OA \perp BC$$ $$AB \cdot AC = AH \cdot AO = AE \cdot AD$$ $$AB \cdot DF = OD \cdot BD$$ and $F$ is midpoint of $DM$.