10. PA and PC are tangents to the circle at A and $$ \mathrm{C} . \mathrm{AD} \| \mathrm{PC} $$, and PD cuts the circle $$ \mathrm{B} . \mathrm{CB} $$ is produced to meet AP at F . $$ \mathrm{AB}, \mathrm{AC} $$ and DC are drawn. Prove: (a) AC bisects PA D (b) $$ \hat{\mathrm{B}}_{1}=\mathrm{B}_{3} $$ (c) $$ \mathrm{APC}=\mathrm{ABD} $$

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$$
\begin{array}{rcl}
\textbf{(a)} &:& \text{Since } PA \text{ and } PC \text{ are tangents from } P \text{ to the circle, we have } PA = PC.\[1mm]
&&\text{So in } \triangle APC,\ \angle P AC = \angle PCA.\[1mm]
&&\text{Because } AD \parallel PC,\ \text{the alternate‐interior angles give}\[1mm]
&&\quad \angle CAD = \angle ACP.\[1mm]
&&\text{But } \angle ACP = \angle PCA \text{ in } \triangle APC, \text{ so } \angle CAD = \angle PCA = \angle P AC.\[2mm]
&&\text{Thus, } AC \text{ bisects } \angle PAD.
\end{array}
$$

\[
\begin{array}{rcl}
\textbf{(b)} &:& \text{In the figure some angles are labelled as } B_1 \text{ and } B_3.\[1mm]
&&\text{A key observation is that the tangent–chord theorem applies at } A \text{ and } C.\[1mm]
&&\text{
(a) AC bisects angle PAD. (b) Angle B₁ equals angle B₃. (c) Angle APC equals angle ABD.

PA and PC are tangents to the
circle at A and , and
PD cuts the circle is
produced to meet AP at F .
and DC are drawn.
Prove:
(a) AC bisects PA D
(b)
©

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PA and PC are tangents to the circle at A and , and PD cuts the circle is produced to meet AP at F . and DC are drawn. Prove: (a) AC bisects PA D (b) ©