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Question Number 67977 by behi83417@gmail.com last updated on 02/Sep/19

Commented by behi83417@gmail.com last updated on 02/Sep/19

$$\mathbf{AD}=\mathbf{DC},\mathrm{\angle }\mathbf{AEB}={90}^{\bullet }.$$$$\mathbf{find}:\mathbf{ED}\mathbf{in}\mathbf{terms}\mathbf{of}:\mathbf{a},\mathbf{b},\mathbf{c}.$$

Answered by $@ty@m123 last updated on 03/Sep/19

$$LetED=x,AE=y\mathrm{\&}BE=h$$$$h^2=c^2-y^2....\left(1\right)$$$$h^2=a^2-{(2x+y)}^2...\left(2\right)$$$$\therefore c^2-y^2=a^2-{(2x+y)}^2$$$$\Rightarrow c^2-y^2=a^2-4x^2-4xy-y^2$$$$4x^2+4xy=a^2-c^2$$$$4x(x+y)=a^2-c^2$$$$4x\times \frac{b}{2}=a^2-c^2$$$$x\times 2b=a^2-c^2$$$$x=\frac{a^2-c^2}{2b}$$

Commented by behi83417@gmail.com last updated on 03/Sep/19

$$\mathrm{right}\mathrm{answer}\mathrm{sir}.\mathrm{thanks}.$$

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