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Question Number 19321 by ajfour last updated on 09/Aug/17

$$\mathrm{Parallel}\mathrm{tangents}\mathrm{to}\mathrm{a}\mathrm{circle}\mathrm{at}\mathrm{A}$$$$\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{cut}\mathrm{in}\mathrm{the}\mathrm{points}\mathrm{C}\mathrm{and}\mathrm{D}$$$$\mathrm{by}\mathrm{a}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{circle}\mathrm{at}\mathrm{E}.$$$$\mathrm{Prove}\mathrm{that}\mathrm{AD},\mathrm{BC}\mathrm{and}\mathrm{the}\mathrm{line}$$$$\mathrm{joining}\mathrm{the}\mathrm{middle}\mathrm{points}\mathrm{of}\mathrm{AE}$$$$\mathrm{and}\mathrm{BE}\mathrm{are}\mathrm{concurrent}.$$

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

$$\mathrm{Let}\mathrm{B}\mathrm{be}\mathrm{origin}.$$$$\mathrm{A}(0,2\mathrm{r}).\mathrm{Let}\mathrm{m}=\tan \theta$$$$\mathrm{C}(\frac{\mathrm{r}}{\mathrm{m}},2\mathrm{r});\mathrm{D}(\mathrm{mr},0)$$$$\mathrm{eq}.\mathrm{of}\mathrm{AD}:\mathrm{y}=2\mathrm{r}-\frac{2}{\mathrm{m}}\mathrm{x}$$$$\mathrm{eq}.\mathrm{of}\mathrm{BC}:\mathrm{y}=2\mathrm{mx}$$$$\mathrm{I}(\alpha ,\beta )\mathrm{lies}\mathrm{on}\mathrm{both},\mathrm{hence}$$$$\beta =2\mathrm{r}-\frac{2\alpha }{\mathrm{m}}=2\alpha \mathrm{m}$$$$\Rightarrow \alpha (\mathrm{m}+\frac{1}{\mathrm{m}})=\mathrm{r}\mathrm{or}\alpha =\frac{\mathrm{mr}}{1+{\mathrm{m}}^2}$$$$\beta =\frac{{2\mathrm{m}}^2\mathrm{r}}{1+{\mathrm{m}}^2}.$$$${\mathrm{x}}_{\mathrm{E}}=\mathrm{rsin}2\theta =\frac{2\mathrm{mr}}{1+{\mathrm{m}}^2}$$$${\mathrm{x}}_{\mathrm{P}}={\mathrm{x}}_{\mathrm{M}}=\frac{{\mathrm{x}}_{\mathrm{E}}}{2}=\frac{\mathbf{mr}}{1+{\mathbf{m}}^2}=\mathit{\alpha }.$$

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