# Choose the correct answer from the given four options:If $\Delta \mathrm{ABC} \sim \Delta \mathrm{EDF}$ and $\Delta \mathrm{ABC}$ is not similar to $\Delta \mathrm{D} \mathrm{EF}$, then which of the following is not true?(A) $\mathrm{BC} \cdot \mathrm{EF}=\mathrm{A} C \cdot \mathrm{FD}$(B) $\mathrm{AB}, \mathrm{EF}=\mathrm{AC} \cdot \mathrm{DE}$(C) $\mathrm{BC} \cdot \mathrm{DE}=\mathrm{AB} \cdot \mathrm{EF}$(D) $\mathrm{BC}, \mathrm{DE}=\mathrm{AB}, \mathrm{FD}$

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Given:

$\Delta \mathrm{ABC} \sim \Delta \mathrm{EDF}$ and $\Delta \mathrm{ABC}$ is not similar to $\Delta \mathrm{D} \mathrm{EF}$

To do:

We have to choose the correct answer.

Solution:

We know that,

If the sides of one triangle are proportional to the side of the other triangle and the corresponding angles are also equal, then the triangles are similar by SSS similarity.

This implies,

$\triangle \mathrm{ABC} \sim \triangle \mathrm{EDF}$

Using similarity property,

$\frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{BC}}{\mathrm{DF}}=\frac{\mathrm{AC}}{\mathrm{EF}}$

This implies,

$\frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{BC}}{\mathrm{DF}}$

$\mathrm{AB} . \mathrm{DF}=\mathrm{ED} . \mathrm{BC}$
$\frac{\mathrm{BC}}{\mathrm{DF}}=\frac{AC}{EF}$

$\mathrm{BC} . \mathrm{EF}=\mathrm{AC} . \mathrm{DF}$

$\frac{\mathrm{AB}}{\mathrm{ED}}=\frac{AC}{EF}$

$\mathrm{AB} . \mathrm{EF}=\mathrm{ED} . \mathrm{AC}$

Therefore,

$\mathrm{BC} \cdot \mathrm{DE}=\mathrm{AB} \cdot \mathrm{EF}$ is not true.

Updated on 10-Oct-2022 13:27:48