Question

Let the circles with centre P and Q touches each other at point A.Let the extended chord AB intersects the circle with centre P at point E and the chord BC touches the circle with centre P at the point D.
Then prove that the ray AD is the angle bisector of $\angle \mathrm{CAE}$.

Answer 1

$$\begin{gathered} \mathrm{Construction}:-\mathrm{Join}\mathrm{ED} \\ \mathrm{Let}\angle \mathrm{ADM}=x\mathrm{and}\angle \mathrm{CAM}=y \\ \mathrm{Since},\mathrm{AM}=\mathrm{DM}(\mathrm{tangents}\mathrm{from}\mathrm{a}\mathrm{external}\mathrm{point}\mathrm{are}\mathrm{equal}) \\ \mathrm{So},\mathrm{AMD}\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{triangle} \\ \angle \mathrm{ADM}=\angle \mathrm{DAM}=x(\mathrm{angles}\mathrm{opposite}toequalsides) \\ \mathrm{Now},\mathrm{In}∆\mathrm{ADC} \\ \angle \mathrm{ACB}=\angle \mathrm{CAD}+\angle \mathrm{ADC}(\mathrm{Exterior}\mathrm{angle}\mathrm{property}) \\ =\angle \mathrm{DAM}+\angle \mathrm{CAM}+\angle \mathrm{ADC} \\ =x+y+x \\ =y+2x \\ \angle \mathrm{CAM}=\angle \mathrm{ABC}=y(\mathrm{angles}\mathrm{in}\mathrm{the}\mathrm{alternate}\mathrm{segment}) \\ \mathrm{Now},\mathrm{In}∆\mathrm{ABC} \\ \angle \mathrm{ABC}+\angle \mathrm{ACB}+\angle \mathrm{CAB}=180° \\ \Rightarrow \angle \mathrm{CAB}=180°-y-y-2x \\ \Rightarrow \angle \mathrm{CAB}=180°-2x-2y...\left(i\right) \\ \angle \mathrm{DAC}=\angle \mathrm{DAM}+\angle \mathrm{MAC}=x+y.....\left(ii\right) \\ \mathrm{Now},\angle \mathrm{EAD}+\angle \mathrm{DAC}+\angle \mathrm{CAB}=180°(\mathrm{EAB}\mathrm{is}\mathrm{a}\mathrm{straight}\mathrm{line}) \\ \Rightarrow \angle \mathrm{EAD}=180°-180°+2x+2y-x-y\left(from\right(i)and(ii\left)\right) \\ \Rightarrow \angle \mathrm{EAD}=x+y \\ \mathrm{Now},\angle \mathrm{EAD}=\angle \mathrm{DAC}=x+y \\ \mathrm{So},\mathrm{AD}\mathrm{is}\mathrm{the}\mathrm{bisector}\mathrm{of}\angle \mathrm{CAE} \end{gathered}$$