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Parallel-tangents-to-a-circle-at-A-and-B-are-cut-in-the-points-C-and-D-by-a-tangent-to-the-circle-at-E-Prove-that-AD-BC-and-the-line-joining-the-middle-points-of-AE-and-BE-are-concurrent-

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Question Number 19321 by ajfour last updated on 09/Aug/17
Parallel tangents to a circle at A and B are cut in the points C and D by a tangent to the circle at E. Prove that AD, BC and the line joining the middle points of AE and BE are concurrent.
$$\mathrm{Parallel}\:\mathrm{tangents}\:\mathrm{to}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{at}\:\mathrm{A} \\ $$$$\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{cut}\:\mathrm{in}\:\mathrm{the}\:\mathrm{points}\:\mathrm{C}\:\mathrm{and}\:\mathrm{D} \\ $$$$\mathrm{by}\:\mathrm{a}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{at}\:\mathrm{E}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{AD},\:\mathrm{BC}\:\mathrm{and}\:\mathrm{the}\:\mathrm{line} \\ $$$$\mathrm{joining}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{points}\:\mathrm{of}\:\mathrm{AE} \\ $$$$\mathrm{and}\:\mathrm{BE}\:\mathrm{are}\:\mathrm{concurrent}. \\ $$
Commented by ajfour last updated on 09/Aug/17
Answered by ajfour last updated on 10/Aug/17
Commented by ajfour last updated on 10/Aug/17
Let B be origin. A(0,2r) . Let m=tan θ C((r/m), 2r) ; D(mr,0) eq. of AD : y=2r−(2/m)x eq. of BC: y=2mx I (α,β) lies on both, hence β=2r−((2α)/m)=2αm ⇒ α(m+(1/m))=r or α=((mr)/(1+m^2 )) β=((2m^2 r)/(1+m^2 )) . x_E =rsin 2θ =((2mr)/(1+m^2 )) x_P =x_M =(x_E /2) = ((mr)/(1+m^2 )) = 𝛂 .
$$\mathrm{Let}\:\mathrm{B}\:\mathrm{be}\:\mathrm{origin}. \\ $$$$\mathrm{A}\left(\mathrm{0},\mathrm{2r}\right)\:\:.\:\mathrm{Let}\:\mathrm{m}=\mathrm{tan}\:\theta \\ $$$$\mathrm{C}\left(\frac{\mathrm{r}}{\mathrm{m}},\:\mathrm{2r}\right)\:\:\:\:;\:\:\mathrm{D}\left(\mathrm{mr},\mathrm{0}\right) \\ $$$$\mathrm{eq}.\:\mathrm{of}\:\mathrm{AD}\::\:\:\:\mathrm{y}=\mathrm{2r}−\frac{\mathrm{2}}{\mathrm{m}}\mathrm{x} \\ $$$$\mathrm{eq}.\:\mathrm{of}\:\mathrm{BC}:\:\:\:\mathrm{y}=\mathrm{2mx} \\ $$$$\mathrm{I}\:\left(\alpha,\beta\right)\:\mathrm{lies}\:\mathrm{on}\:\mathrm{both},\:\mathrm{hence} \\ $$$$\:\:\beta=\mathrm{2r}−\frac{\mathrm{2}\alpha}{\mathrm{m}}=\mathrm{2}\alpha\mathrm{m} \\ $$$$\Rightarrow\:\:\alpha\left(\mathrm{m}+\frac{\mathrm{1}}{\mathrm{m}}\right)=\mathrm{r}\:\:\mathrm{or}\:\:\alpha=\frac{\mathrm{mr}}{\mathrm{1}+\mathrm{m}^{\mathrm{2}} }\: \\ $$$$\:\:\:\:\:\beta=\frac{\mathrm{2m}^{\mathrm{2}} \mathrm{r}}{\mathrm{1}+\mathrm{m}^{\mathrm{2}} }\:. \\ $$$$\mathrm{x}_{\mathrm{E}} =\mathrm{rsin}\:\mathrm{2}\theta\:=\frac{\mathrm{2mr}}{\mathrm{1}+\mathrm{m}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\mathrm{x}_{\mathrm{P}} =\mathrm{x}_{\mathrm{M}} =\frac{\mathrm{x}_{\mathrm{E}} }{\mathrm{2}}\:=\:\frac{\boldsymbol{\mathrm{mr}}}{\mathrm{1}+\boldsymbol{\mathrm{m}}^{\mathrm{2}} }\:=\:\boldsymbol{\alpha}\:. \\ $$

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