Question

prove that diagonal of a rectangle divide it into two triangle congruent to each other

Answer 1

here, in rectangle ABDC,
AD is a diagonal
so,
in ∆ABD and ∆DCA
AB=DC ( opposite sides of rectangle )
BD=CA ( opposite sides of rectangle )
AD=DA ( common )
so,
∆ABD is congruent to ∆DCA
by SSS

Answer 2

Proved that  diagonal of a Rectangle divide it into two triangle congruent to each other

Step-by-step explanation:

Let say ABCD is rectangle

rectangle has all the angles = 90°

=> ∠A = ∠B = ∠C = ∠D = 90°

opposite sides of rectangle are equal

=> AB = CD

& BC = AD

let say Diagonal AC

divides rectangle ABCD in

ΔABC  & CDA

AB = CD

∠B  = ∠D = 90°

AC = AC  ( common)

BC = AD

=> ΔABC ≅ CDA

Now let say Diagonal BD

divides rectangle ABCD in

ΔBAD  & DCB

AB = CD

∠A  = ∠C = 90°

AC = AC  ( common)

AD = BC

=> ΔBAD  ≅ DCB

Hence proved diagonal of a Rectangle divide it into two triangle congruent to each other

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