Question

In $△DFH$, point $E$ is the midpoint of hypotenuse $DF$ segment $HE$ is perpendicular to the diagonal $DF$, segment $EG$ is perpendicular to segment $FH$,segment $EK$ is perpendicular to segment $DH$.Show that ${EG}^2=FG\times EK$

Answer 1

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Given : $\angle G={90}^{\circ }=\angle K=\angle H=\angle DEH$

$\therefore EGHK$ is a square

EF=ED

To prove : $E{G}^2=FG\times EK$

in $\Delta$ EFH & EHD

EF=ED (Given) _______ (1)

EH $\Rightarrow$ common sie ________ (2)

$\angle DEH=\angle FEH={90}^{\circ }$

$\therefore \Delta DEH\cong \Delta FEH$ (SAS congruency)

$\Rightarrow FH=DH$

$\Rightarrow \angle DFG=\angle FDG={45}^{\circ }$

In $\Delta FGE$

tan < EFG = $\frac{EG}{FG}$

$\Rightarrow tan45=\frac{EG}{FG}$

$\Rightarrow EG=FG$ _______ (3)

EG=EK [ EGHK is square]________ (4)

from (3) & (4)

$(EG{)}^2=FG\times EK$