In the given figure,


In the given figure,

(i) $\triangle \mathrm{DMU} \sim \triangle \mathrm{BMV}$


(ii) $\mathrm{DM} \times \mathrm{BV}=\mathrm{BM} \times \mathrm{DU}$


(i) Given $A B \| D C$

$\angle M U D=\angle M V B$

Each angle is equal to $90^{\circ}$

$\angle U M D=\angle V M B$

Each are vertically opposite angles.

Therefore, by AA-criterion of similarity $\triangle D M U \sim \triangle B M V$

(ii)Since $\triangle D M U \sim \triangle B M V$

$\frac{D M}{B M}=\frac{M U}{M V}=\frac{D U}{B V}$

$\frac{D M}{B M}=\frac{D U}{B V}$

By cross multiplication, we get $D M \times B V=D U \times B M$

Hence proved that $D M \times B V=D U \times B M$

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