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Question-49362




Question Number 49362 by behi83417@gmail.com last updated on 06/Dec/18
Commented by behi83417@gmail.com last updated on 06/Dec/18
as showen in picture,when this statement  is true?                FE^2 +FD^2 =BC^2
$$\boldsymbol{\mathrm{as}}\:\boldsymbol{\mathrm{showen}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{picture}},\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{statement}} \\ $$$$\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{true}}? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{FE}}^{\mathrm{2}} +\boldsymbol{\mathrm{FD}}^{\mathrm{2}} =\boldsymbol{\mathrm{BC}}^{\mathrm{2}} \\ $$
Commented by mr W last updated on 06/Dec/18
when AE=AD and F=H, i.e. ED  tangents the circle.
$${when}\:{AE}={AD}\:{and}\:{F}={H},\:{i}.{e}.\:{ED} \\ $$$${tangents}\:{the}\:{circle}. \\ $$
Commented by behi83417@gmail.com last updated on 06/Dec/18
this is true when ED∥BC,even if:F≠H.  right sir?
$${this}\:{is}\:{true}\:{when}\:{ED}\parallel{BC},{even}\:{if}:{F}\neq{H}. \\ $$$${right}\:{sir}? \\ $$
Commented by mr W last updated on 06/Dec/18
no. this is true only ED//BC and F=H.
$${no}.\:{this}\:{is}\:{true}\:{only}\:{ED}//{BC}\:{and}\:{F}={H}. \\ $$
Commented by mr W last updated on 06/Dec/18
I misread the question as  FE^2 +HD^2 =BC^2
$${I}\:{misread}\:{the}\:{question}\:{as} \\ $$$${FE}^{\mathrm{2}} +{HD}^{\mathrm{2}} ={BC}^{\mathrm{2}} \\ $$
Commented by behi83417@gmail.com last updated on 06/Dec/18
yes sir.is it ⇑true in special case?
$${yes}\:{sir}.{is}\:{it}\:\Uparrow{true}\:{in}\:{special}\:{case}? \\ $$
Commented by mr W last updated on 06/Dec/18
yes. there are also infinite possibilities.
$${yes}.\:{there}\:{are}\:{also}\:{infinite}\:{possibilities}. \\ $$
Answered by behi83417@gmail.com last updated on 06/Dec/18
Commented by behi83417@gmail.com last updated on 06/Dec/18
∡HDB=∡GEK=45  ∡HGI=∡KGE=45  DG^2 =DI^2 +IG^2 =2IG^2   GE^2 =GK^2 +KE^2 =2KG^2   ⇒DG^2 +GE^2 =2(IG^2 +KG^2 )=2AG^2 =  =2R^2 =BC^2  .
$$\measuredangle{HDB}=\measuredangle{GEK}=\mathrm{45} \\ $$$$\measuredangle{HGI}=\measuredangle{KGE}=\mathrm{45} \\ $$$${DG}^{\mathrm{2}} ={DI}^{\mathrm{2}} +{IG}^{\mathrm{2}} =\mathrm{2}{IG}^{\mathrm{2}} \\ $$$${GE}^{\mathrm{2}} ={GK}^{\mathrm{2}} +{KE}^{\mathrm{2}} =\mathrm{2}{KG}^{\mathrm{2}} \\ $$$$\Rightarrow{DG}^{\mathrm{2}} +{GE}^{\mathrm{2}} =\mathrm{2}\left({IG}^{\mathrm{2}} +{KG}^{\mathrm{2}} \right)=\mathrm{2}{AG}^{\mathrm{2}} = \\ $$$$=\mathrm{2}{R}^{\mathrm{2}} ={BC}^{\mathrm{2}} \:. \\ $$

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