Question

Two circles with centres P and Q touch each other at point T externally, seg BD is a diameter of the circle with centre Q,line BA is a common tangent touching the circle at A.Prove that the points D,T and A are collinear.

Answer 1

$$\begin{gathered} \angle \mathrm{BTD}=90°(\mathrm{angle}\mathrm{in}\mathrm{the}\mathrm{semicircle}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{angle}) \\ \angle \mathrm{ABD}=90°(\mathrm{radius}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{tangent}) \\ \mathrm{AB}\mathrm{is}\mathrm{tangent}\mathrm{and}\mathrm{AD}\mathrm{is}\mathrm{secant} \\ \mathrm{So}\mathrm{by}\mathrm{tangent}\mathrm{secant}\mathrm{property} \\ {\mathrm{AB}}^2=\mathrm{AT}\times \mathrm{AD} \\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AT}}{\mathrm{AB}} \\ \angle \mathrm{A}=\angle \mathrm{A}\left(\mathrm{common}\right) \\ \mathrm{By}\mathrm{SAS}\mathrm{similarity}∆\mathrm{ATB}~∆\mathrm{ABD} \\ \mathrm{so},\angle \mathrm{ATB}=\angle \mathrm{ABD}=90° \\ \mathrm{Hence},\angle \mathrm{BTD}+\angle \mathrm{ATB}=180° \\ \mathrm{So}\mathrm{D},\mathrm{T},\mathrm{A}\mathrm{are}\mathrm{collinear}. \end{gathered}$$