In trapezium square ABCD,overline{AB} paralle | vedclass

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(N/A) $1$. Given: In trapezium $\square ABCD$,$\overline{AB} \parallel \overline{CD}$. The diagonals $\overline{AC}$ and $\overline{BD}$ intersect at

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(N/A) $1$. Given: In trapezium $\square ABCD$,$\overline{AB} \parallel \overline{CD}$. The diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $M$.
$2$. Consider $\Delta MAB$ and $\Delta MCD$.
$3$. Since $\overline{AB} \parallel \overline{CD}$,the alternate interior angles formed by the transversal lines $\overline{AC}$ and $\overline{BD}$ are equal.
$4$. Therefore,$\angle MAB = \angle MCD$ (alternate interior angles).
$5$. Similarly,$\angle MBA = \angle MDC$ (alternate interior angles).
$6$. Also,$\angle AMB = \angle CMD$ (vertically opposite angles).
$7$. By the $AAA$ (Angle-Angle-Angle) similarity criterion,$\Delta MAB \sim \Delta MCD$.
$8$. Thus,the correspondence $MAB \leftrightarrow MCD$ is a similarity.

In trapezium $\square ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{M\}$. Prove that the correspondence $MAB \leftrightarrow MCD$ between $\Delta MAB$ and $\Delta MCD$ is a similarity.

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