Math, asked by Anonymous, 4 months ago

if the diagonals of a parallelogram are equal then show that it is a rectangle .

Answers

Answered by BadCaption01
2

\huge\bold{Given :-}

  • Parallelogram ABCD , AC = BD

\huge\bold{To~prove :-}

  • ABCD is a rectangle.

\huge\bold{Proof :-}

In \triangle ABC and \triangle DCB

AB = CD \bf\blue{(opposite\:sides\:of\:||\:gm\:are\:equal)} \\

AC = BD \bf\red{(Given)} \\

BC = BC \bf\green{(Common)} \\

\ast \triangle ABC = \triangle DCB

\Rightarrow By SSS congruence criteria

\ast ∠1 = ∠2 \bf\orange{(by~cpct)} \\

AB || CD\bf\red{(opposite~sides~of~||~gm~are~parallel.)} \\

And \bf\red{( BC~is~the~transversal)} \\

\ast ∠1 = ∠2 = 180° \bf\purple{(co~-interior~angles)} \\

\Rightarrow 2∠1 = 180° -------- \bf\pink{(using~(1))} \\

\Rightarrow ∠1 = 90°

\ast ∠1 = ∠2 = 90° -------\bf\pink{(2)} \\

Also, opposite angles of parallelogram are equal.

\ast ∠2 = ∠D = 90°

∠1 = ∠A = 90°

All the angles of parallelogram ABCD are 90°

\Rightarrow \boxed {\sf {\purple { ABCD~is~a~rectangle}}}

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