Question

ABCD IS A SQUARE E F AND G ARE MIDPOINTS OF SIDE AB BC CD RESPECTIELY PROVE THAT THE TRINGLES AEF AND DGF ARE CONGRUENT

Answer 1

$\bold{ \: \underline{In \: \: \square \: \: ABCD , \: \: all \: \: sides \: \: will \: \: be \: \: equal \: \: as \: \: it \: \: is \: \: a \: \: square. }}$



$\bold{ \: So, AB = BC = CD = DA }$





And ,
$\bold{E \: is \: the \: mid \: point \: of \: AB} \\ \bold{ F \: is \: the \: mid \: point \: of \: BC} \\ \bold{ G \: is \: the \: mid \: point \: of \: CD}$


$\bold{So, } \\ \\ \bold{ AE = EB = \frac{AB}{2} \: \: \: \: \: \: \: \: \: \: \: \: ...(i)} \\ \\ \bold{ BF = FC = \frac{ BC}{2} \: \: \: \: \: \: \: \: \: \: \: ...(ii)} \\ \\ \bold{ CG = GD = \frac{ CD }{2} \: \: \: \: \: \: \: \: \: ...(iii)}$




And,


$\bold{ We \: know, all \: angles \: of \: a \: Square \: are \: 90°}$







In ∆ BEF,

$\bold{Hence,By \: P ythagoras \: Theorem, } \\ \\ \bold{ \: EB^{2} + BF^{2} = EF^{2}} \: \: \: \: \: ..(iv)$



In ∆ GFC,

$\bold{GC^{2} +CF^{2} = GF^{2} } \\ \\ \\ \bold{from \: (i) \: and \: (ii)} \\ \\ \\ \\ \\ \bold{{( \frac{CD}{2} )}^{2} + {( \frac{BC }{2}) }^{2} = {GF}^{2} }\\ \\ \\ \bold{{( \frac{ AB }{2}) }^{2} + {( \frac{BC }{2} )}^{2} = { GF}^{2} } \\ \\ \\ \bold{{E B}^{2} + {BF}^{2} = {GF}^{2} } \: \: \: \: \: \: \: \: ...(v)$





$\bold{We \: know, sides \: are \: equal, }$

$\bold{AB = CD} \\ \\ \bold { Divide \: both \: sides \: by \: 2} \: \\ \\ \\ \\ \frac{AB}{2} = \frac{ CD }{2} \\ \\ \\ \bold{AE = EB = GD = GC } \\ \\ \\ \bold{ AE = GD} \: \: \: \: \: \: \: \: \: ...(vi)$





In ∆ ABF,

$\bold{Now, Again \: by \: Pythagoras \: theorem, } \\ \\ \bold{ AB^{2} + BF^{2} = AF^{2}} \: \: \: \: \: \: \: \: \: ...(vii)$





In ∆ DFC,

$\bold{CD^{2} + CF^{2} =DF^{2} } \\ \\ \bold{ AB^{2} + BF^{2} = DF^{2} \: \: \: \: \: ...(viii) \: \: \: \: \: \: \: \: \: \: [ All \: sides \: are \: equal \: so \: CD = AD \: and \: CF = \frac{ BC}{2}= BF] }$









Hence,


• ( iv ) = ( v )
EF² = GF²
EF = GF



• AE = GD [ From ( vi ) ]



• ( vii ) =( viii )
AF² = DF²
AF = DF





Hence,
$\bold{AEF \cong \: DGF \: \: By \: \: S.S.S }$

Answer 2

THIS is proved above.

MAY THIS WILL HELP YOU A LOT

THANX...