Question

If the diagonals of a parallelogram are equal, then show that it is a rectangle​

Answer 1

$\huge \boxed{ \red{question}}$

If the diagonals of a parallelogram are equal, then show that it is a rectangle

$\huge \pink{ \to find}$

ABCD is a rectangle..

$\boxed{ \purple{given}}$

ABCD is a parallelogram

$\huge \red{ \{solution} \}$

consider Δ ACD and Δ ABD

AC = BD .... (given)

AB = DC .... (opposite sides of parallelogram)

AD = AD .... (common side)

∴Δ ACD ≅Δ ABD (sss test of congruence)

∠ BAD = ∠ CDA .... (cpct)

∠BAD+∠CDA=180 [Adjacent angles of parallelogram are supplementary]

so ∠ BAD and ∠ CDA are right angles as they are congruent and supplementary.

Therefore, □ ABCD is a rectangle since a

parallelogram with one right interior angle is a rectangle.

Answer 2

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$\sf\underbrace{Appropriate\:Question: }$

• If the diagonals of a parallelogram are equal, then show that it is a rectanglel?

$\sf\underbrace{Required\: Answer: }$

$\bf\underline{Given: }$

• Let ABCD be a parallelogram Where AC = BC

$\bf\underline{To\:prove:}$

• ABCD is a rectangle

$\bf\underline{Solution:}$

Rectangle is a parallelogram with one angle 90° We prove that one of interior angles is 90°.

in ∆ABC and ∆DCB,

$\implies$$\sf{AB\:=\:DC}$

$\implies$$\sf{BC\:=\:BC}$

$\implies$$\sf{AC\:=\:DB}$

$\implies$$\sf{ ∴\:∆ABC\:≈\:∆DCB}$

$\implies$$\sf{ ∠ABC\:≈\:∠DCB}$

Now

AB || DC & BC is a transversal

$\implies$$\sf{∴\:∠B\:∠C\:=\:180°}$

$\implies$$\sf{∠B\:∠C\:=\:180°}$

$\implies$$\sf{2\:∠B\:=\:180°}$

$\implies$∠B = $\sf\dfrac{180°}{2}$ = 90°

So, ABCD is a parallelogram with one angle 90°

$\bf\underline{∴ \:ABCD\: is \:a \:rectangle}$

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