Question Number 67977 by behi83417@gmail.com last updated on 02/Sep/19
Commented by behi83417@gmail.com last updated on 02/Sep/19
$$\boldsymbol{\mathrm{AD}}=\boldsymbol{\mathrm{DC}},\angle\boldsymbol{\mathrm{AEB}}=\mathrm{90}^{\bullet} \:\:. \\ $$$$\boldsymbol{\mathrm{find}}:\:\:\:\boldsymbol{\mathrm{ED}}\:\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{terms}}\:\boldsymbol{\mathrm{of}}:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}. \\ $$
Answered by $@ty@m123 last updated on 03/Sep/19
$${Let}\:{ED}={x},\:{AE}={y}\:\&\:{BE}={h} \\ $$$${h}^{\mathrm{2}} ={c}^{\mathrm{2}} −{y}^{\mathrm{2}} \:\:....\left(\mathrm{1}\right) \\ $$$${h}^{\mathrm{2}} ={a}^{\mathrm{2}} −\left(\mathrm{2}{x}+{y}\right)^{\mathrm{2}} \:...\left(\mathrm{2}\right) \\ $$$$\therefore\:{c}^{\mathrm{2}} −{y}^{\mathrm{2}} \:={a}^{\mathrm{2}} −\left(\mathrm{2}{x}+{y}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{c}^{\mathrm{2}} −{y}^{\mathrm{2}} \:={a}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{xy}−{y}^{\mathrm{2}} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{xy}={a}^{\mathrm{2}} −{c}^{\mathrm{2}} \\ $$$$\mathrm{4}{x}\left({x}+{y}\right)={a}^{\mathrm{2}} −{c}^{\mathrm{2}} \\ $$$$\mathrm{4}{x}×\frac{{b}}{\mathrm{2}}={a}^{\mathrm{2}} −{c}^{\mathrm{2}} \\ $$$${x}×\mathrm{2}{b}={a}^{\mathrm{2}} −{c}^{\mathrm{2}} \\ $$$${x}=\frac{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{b}} \\ $$
Commented by behi83417@gmail.com last updated on 03/Sep/19
$$\mathrm{right}\:\mathrm{answer}\:\mathrm{sir}.\mathrm{thanks}. \\ $$