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Question-210014




Question Number 210014 by Ismoiljon_008 last updated on 28/Jul/24
Commented by Ismoiljon_008 last updated on 28/Jul/24
   Help Please
$$\:\:\:\mathscr{H}{elp}\:\mathscr{P}{lease} \\ $$
Commented by a.lgnaoui last updated on 28/Jul/24
∡FBA=∡FBC?      AB//CD?
$$\measuredangle\mathrm{FBA}=\measuredangle\mathrm{FBC}?\:\:\:\:\:\:\mathrm{AB}//\mathrm{CD}? \\ $$
Commented by Ismoiljon_008 last updated on 28/Jul/24
   Yes,  ABCD is parallelogram and     BF  is bissector.
$$\:\:\:\mathscr{Y}{es},\:\:{ABCD}\:{is}\:{parallelogram}\:{and} \\ $$$$\:\:\:{BF}\:\:{is}\:{bissector}. \\ $$
Answered by A5T last updated on 28/Jul/24
Commented by A5T last updated on 28/Jul/24
Through F, construct a line parallel to AB.  ⇒((AG)/(AD))=((AF)/(AE))=((FG)/(DE))=(1/2)⇒AG=GD=(x/2);FG=2.5  ⇒(x/2)+2.5=10⇒x=15
$${Through}\:{F},\:{construct}\:{a}\:{line}\:{parallel}\:{to}\:{AB}. \\ $$$$\Rightarrow\frac{{AG}}{{AD}}=\frac{{AF}}{{AE}}=\frac{{FG}}{{DE}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{AG}={GD}=\frac{{x}}{\mathrm{2}};{FG}=\mathrm{2}.\mathrm{5} \\ $$$$\Rightarrow\frac{{x}}{\mathrm{2}}+\mathrm{2}.\mathrm{5}=\mathrm{10}\Rightarrow{x}=\mathrm{15} \\ $$
Commented by Ismoiljon_008 last updated on 29/Jul/24
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$
Answered by mr W last updated on 28/Jul/24
Commented by mr W last updated on 28/Jul/24
CG=AD=x  EG=AE ⇒FG=3×AF  ((10)/(AF))=((2x)/(FG)) ⇒10=((2x)/3) ⇒x=15
$${CG}={AD}={x} \\ $$$${EG}={AE}\:\Rightarrow{FG}=\mathrm{3}×{AF} \\ $$$$\frac{\mathrm{10}}{{AF}}=\frac{\mathrm{2}{x}}{{FG}}\:\Rightarrow\mathrm{10}=\frac{\mathrm{2}{x}}{\mathrm{3}}\:\Rightarrow{x}=\mathrm{15} \\ $$
Commented by Ismoiljon_008 last updated on 29/Jul/24
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$

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