square ABCD is a trapezium in which overline{ | vedclass

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(A) Given that $\square ABCD$ is a trapezium with $\overline{AB} \parallel \overline{CD}$.We are given $AM 	imes NC = BN 	imes MD$,which can be rewr

Vedclass · Accepted Answer

$\frac{2}{5}$

Vedclass · Answer

(A) Given that $\square ABCD$ is a trapezium with $\overline{AB} \parallel \overline{CD}$.We are given $AM 	imes NC = BN 	imes MD$,which can be rewritten as $\frac{AM}{MD} = \frac{BN}{NC}$.Since $\frac{AM}{MD} = \frac{2}{3}$,it follows that $\frac{BN}{NC} = \frac{2}{3}$.By the Basic Proportionality Theorem (Thales Theorem) or its converse,since $\frac{AM}{MD} = \frac{BN}{NC}$,the line segment $\overline{MN}$ must be parallel to the bases $\overline{AB}$ and $\overline{CD}$.Let $\overline{MN} \parallel \overline{AB} \parallel \overline{CD}$.In $	riangle ADC$,since $\overline{MO} \parallel \overline{CD}$,by the intercept theorem (or similarity of triangles $	riangle AMO \sim 	riangle ADC$),we have $\frac{AO}{AC} = \frac{AM}{AD}$.Given $\frac{AM}{MD} = \frac{2}{3}$,let $AM = 2k$ and $MD = 3k$. Then $AD = AM + MD = 2k + 3k = 5k$.Therefore,$\frac{AO}{AC} = \frac{AM}{AD} = \frac{2k}{5k} = \frac{2}{5}$.

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