In trapezium ABCD- M -AD- n -BC are the points such that AMMD - BNNC - 23-The diagonal AC intersects MN at O- Then find the value of AOAC-

In □ ABCD, AB || CD , M and N are the points on transversals AD and BC respectivelyAlso, AMMD = BNNC ∴MN || AB and ...

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Standard VII

Mathematics

Angle of Rotational Symmetry

In trapezium ...

Question

In trapezium ABCD, $M \in ¯ ¯¯¯¯¯¯¯ ¯ A D, n \in ¯ ¯¯¯¯¯¯ ¯ B C$ are the points such that $\frac{A M}{M D} = \frac{B N}{N C} = \frac{2}{3}$ .
The diagonal $¯ ¯¯¯¯¯¯ ¯ A C$ intersects $¯ ¯¯¯¯¯¯¯¯¯ ¯ M N$ at $O$ . Then find the value of $\frac{A O}{A C}$ .

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Solution

In $□ A B C D, ¯ ¯¯¯¯¯¯ ¯ A B | | ¯ ¯¯¯¯¯¯¯ ¯ C D$ , $M$ and $N$ are the points on transversals $\leftarrow \to A D$ and $\leftarrow \to B C$ respectively
Also, $\frac{A M}{M D} = \frac{B N}{N C}$
$∴ \leftarrow - \to M N | | \leftarrow \to A B$ and $\leftarrow \to A B | | \leftarrow \to C D$
$\leftarrow - \to M N$ intersects $\leftarrow \to A C$ at $O$ .
$∴$ In $Δ A D C, \leftarrow - \to M O | | \leftarrow \to D C, M ε ¯ ¯¯¯¯¯¯¯ ¯ A D, O ε ¯ ¯¯¯¯¯¯ ¯ A C$
$∴ \frac{A M}{M D} = \frac{A O}{O C}$
But $\frac{A M}{M D} = \frac{2}{3}$
$∴ \frac{A O}{O C} = \frac{2}{3}$
$∴ \frac{A O}{A O + O C} = \frac{2}{2 + 3}$
$∴ \frac{A O}{A C} = \frac{2}{5}$

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In trapezium ABCD, M∈¯¯¯¯¯¯¯¯¯AD,n∈¯¯¯¯¯¯¯¯BC are the points such that AMMD=BNNC=23.The diagonal ¯¯¯¯¯¯¯¯AC intersects ¯¯¯¯¯¯¯¯¯¯¯MN at O. Then find the value of AOAC.

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The diagonal $¯ ¯¯¯¯¯¯ ¯ A C$ intersects $¯ ¯¯¯¯¯¯¯¯¯ ¯ M N$ at $O$ . Then find the value of $\frac{A O}{A C}$ .