1. **Problem Statement:** Given a circle with center O and diameter AB, OD is parallel to AC, and EF is the tangent at D. Angle $B\hat{C}D = x^\circ$. We need to find angles $B\hat{O}D$, $C\hat{A}O$, and $B\hat{D}C$ in terms of $x$, and prove that (a) $CD = BD$, (b) $EF \parallel BC$, and (c) $A\hat{C}B = E\hat{D}O$. 2. **Known facts and formulas:** - Angle subtended by diameter is $90^\circ$. - Tangent to a circle is perpendicular to radius at point of contact. - Alternate interior angles are equal if lines are parallel. - In a circle, equal chords subtend equal angles at the center. 3. **Step i: Marking data on figure** - $AB$ is diameter. - $OD \parallel AC$. - $EF$ tangent at $D$. - $B\hat{C}D = x^\circ$. 4. **Step ii: Find angles in terms of $x$** - Since $AB$ is diameter, $A\hat{B}D = 90^\circ$. - $OD \parallel AC$ implies $B\hat{O}D = B\hat{C}D = x^\circ$ (alternate interior angles). Thus, $$B\hat{O}D = x^\circ$$ - Triangle $AOC$ is isosceles with $OA = OC$ (radii), so angles at $A$ and $C$ are equal. - Since $OD \parallel AC$, angle $C\hat{A}O = B\hat{C}D = x^\circ$. Thus, $$C\hat{A}O = x^\circ$$ - Angle $B\hat{D}C$ is equal to $B\hat{C}D$ because $BD = CD$ (to be proved), so $$B\hat{D}C = x^\circ$$ 5. **Step iii: Prove the statements** (a) **$CD = BD$** - In triangle $BCD$, angles $B\hat{C}D = x^\circ$ and $B\hat{D}C$ are equal. - By the Isosceles triangle property, sides opposite equal angles are equal. - Therefore, $$CD = BD$$ (b) **$EF \parallel BC$** - $EF$ is tangent at $D$, so $OD \perp EF$. - Since $OD \parallel AC$ and $AC$ intersects $BC$, by alternate interior angles, - $EF$ is parallel to $BC$. (c) **$A\hat{C}B = E\hat{D}O$** - $A\hat{C}B$ is an angle subtended by chord $AB$ at $C$. - $E\hat{D}O$ is angle between tangent $EF$ and radius $OD$. - By tangent-secant theorem, these angles are equal. **Final answers:** $$B\hat{O}D = x^\circ, \quad C\hat{A}O = x^\circ, \quad B\hat{D}C = x^\circ$$ $$CD = BD, \quad EF \parallel BC, \quad A\hat{C}B = E\hat{D}O$$