In the given figure, AB $∥$ DC Prove that(i) $△ D M U \sim △ B M V$ (ii) $D M \times B V = B M \times D U$

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Solution

In $△ D M U a n d △ B M V$

$(i)$ Since $A B ∥ D C$ AND $D B$ is transversal,

$∠ U D M = ∠ M B V$ (alternate angles)

Since $A B ∥ D C$ AND $U V$ is transversal,

$∠ D U M = ∠ M V B$ (alternate angles)

$∠ D M U = ∠ V M B$ (vertically opposite angles)

$∴ △ D M U \sim △ B M V$ (AAA similarity criterion)
Hence, proved.

$(i i)$ Since $△ D M U \sim △ B M V$ ,

By CPST,
$\frac{D M}{B M} = \frac{D U}{B V}$

$∴ D M \times B V = D U \times B M$
Hence, proved.

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In the given figure, AB ∥ DC Prove that(i) △DMU∼△BMV(ii) DM×BV=BM×DU

In the given figure, AB $∥$ DC Prove that(i) $△ D M U \sim △ B M V$ (ii) $D M \times B V = B M \times D U$