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Prove that the diagonals of a parallelogram bisect each other​

User Ndori
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Answer:

[ See the attached picture ]

The diagonals of a parallelogram bisect each other.

✧ Given : ABCD is a parallelogram. Diagonals AC and BD intersect at O.

✺ To prove : AC and BD bisect each other at O , i.e AO = OC and BO = OD.

Proof :
\begin{array} c  \hline \tt{SN}&amp; \tt{STATEMENTS} &amp; \tt{REASONS}\\ \hline 1&amp; \sf{In \: \triangle ^(s) \:AOB \: and \: COD } \\ \sf{(i)}&amp; \sf{ \angle \: OAB = \angle \: OCD\: (A)}&amp; \sf{AB \parallel \: DC \: and \: alternate \: angles} \\ \sf{(ii)} &amp;\sf{AB = DC(S)}&amp; \sf{Opposite \: sides \: of \: a \: parallelogram} \\ \sf{(iii)} &amp;\sf{ \angle \: OBA= \angle \: ODC(A)} &amp;\sf{AB \parallel \:DC \: and \: alternate \: angles} \\ \sf{(iv)}&amp; \sf{ \triangle \:AOB\cong \triangle \: COD}&amp; \sf{A.S.A \: axiom}\\ \hline 2.&amp; \sf{AO = OC \: and \: BO = OD}&amp; \sf{Corresponding \: sides \: of \: congruent \: triangle}\\ \hline 3.&amp; \sf{AC \: and \: BD \: bisect \: each \: other \: at \: O}&amp; \sf{From \: statement \: (2)}\\ \\ \hline\end{array}. </p><p>Proved ✔ </p><p></p><p>

♕ And we're done! Hurrayyy! ;)

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Prove that the diagonals of a parallelogram bisect each other​-example-1
User Strange Quark
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