Begin by joining the point O to the point A and marking all the radiuses as equal lengths on the diagram.

0XZ7dlRs_4-8-edexcel-gcse-vh-q2-diagram-1

This creates two isosceles triangles, triangle AOB and triangle AOC.
The base angles of isosceles triangles are equal, mark these equal angles on the diagram, labelling the angles in one triangle with one variable and the angles in the other triangle with a different variable.

Let angle $O A B = x$ and angle $O A C = y$ .
Then angle $B A C = x + y$ .

$O \hat{A} B = O \hat{B} A = x$

$O \hat{A} C = O \hat{C} A = y$

4-8-edexcel-gcse-vh-q2-diagram-2

Write angle BAC in terms of $x$ and $y$ .

$B \hat{A} C = x + y$

[1]

Angles in a triangle add up to 180°, use this to find the angles AOC and BOC.

$A \hat{O} B = 180 - x - x = 180 - 2 x$

$A \hat{O} C = 180 - y - y = 180 - 2 y$

Angles at a point add up to 360°, use this to find an expression for angle BOC in terms of $x$ and $y$ .

$B \hat{O} C + A \hat{O} B + A \hat{O} C = 360 ° B \hat{O} C + (180 - 2 x) + (180 - 2 y) = 360$

[1]

Simplify.

$\begin{array}{rcl} B \hat{O} C + 180 - 2 x + 180 - 2 y & = & 360 \\ B \hat{O} C + 360 - 2 x - 2 y & = & 360 \\ B \hat{O} C - 2 (x + y) & = & 0 \\ B \hat{O} C & = & 2 (x + y) \end{array}$

[1]

We have already shown that $B \hat{A} C = x + y$ .

$\begin{array}{rcl} B \hat{O} C & = & 2 (x + y) = 2 B \hat{A} C \end{array}$

The base angles of an isosceles triangle are equal
Angles at a point add up to 360°

Therefore Angle BOC = 2(Angle BAC )All reasons given clearly [1]

$□$	$A B$ is parallel to $D C$
$□$	$A C$ bisects angle $B C D$
$□$	$A C$ bisects angle $B A D$

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Circle Theorems