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Question Number 139649 by mnjuly1970 last updated on 30/Apr/21

Commented by mr W last updated on 30/Apr/21

$$\frac{red}{green}=\frac{6}{5}$$

Answered by mr W last updated on 30/Apr/21

$$DE//AB(\ast )$$$$ABistangentatF.$$$$$$$$(CE+2)\times 2=4^2(\ast \ast )$$$$\Rightarrow CE=6$$$$\frac{DE}{AB}=\frac{CE}{CB}=\frac{6}{6+2}=\frac{6}{8}$$$$\Rightarrow DE=\frac{6\times (6+4)}{8}=\frac{15}{2}$$$$\frac{AC}{DC}=\frac{CB}{CE}=\frac{8}{6}=\frac{4}{3}$$$$AC=\frac{4}{3}\times DC$$$$AD\times AC=6^2(\ast \ast )$$$$(AC-DC)\times AC=6^2$$$$(\frac{4}{3}\times DC-DC)\times \frac{4}{3}\times DC=6^2$$$$\frac{4}{9}\times {DC}^2=6^2$$$$\Rightarrow DC=\frac{6\times 3}{2}=9$$$$\Rightarrow \frac{DC}{DE}=\frac{9\times 2}{15}=\frac{6}{5}$$

Commented by mr W last updated on 30/Apr/21

Commented by mnjuly1970 last updated on 30/Apr/21

$$verynicethanksalotmrW..$$$$grateful..$$

Commented by mr W last updated on 30/Apr/21

Commented by mr W last updated on 30/Apr/21

$$prooffor(\ast )$$$$CGisthecommontangentofboth$$$$circlesatpointC.$$$$\mathrm{\angle }CDE=\mathrm{\angle }GCE=\alpha$$$$\mathrm{\angle }CAB=\mathrm{\angle }GCB=\alpha$$$$\Rightarrow \mathrm{\angle }CDE=\mathrm{\angle }CAB$$$$\Rightarrow DE//AB$$

Commented by mr W last updated on 30/Apr/21

$$prooffor(\ast \ast )$$$$seeQ137946$$

Commented by mnjuly1970 last updated on 30/Apr/21

$$thanksalot...$$

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