In the following figure, DE || AC and DC || AP. Prove that : BE/EC = BC/CP.

Given, DE || AC In △BAC, By basic proportionality theorem, .....(1) Given, DC || AP In △BPA, By basic proportionality theorem, .....(2) . Hence, proved that .

Mathematics

In the following figure, DE || AC and DC || AP. Prove that : $\dfrac{BE}{EC} = \dfrac{BC}{CP}$ .

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Given, DE || AC

In △BAC,

By basic proportionality theorem,

$\dfrac{BE}{EC} = \dfrac{BD}{AD}$ …..(1)

Given, DC || AP

In △BPA,

By basic proportionality theorem,

$\dfrac{BC}{CP} = \dfrac{BD}{DA}$ …..(2)

$\dfrac{BE}{EC} = \dfrac{BC}{CP}$ .

Hence, proved that $\dfrac{BE}{EC} = \dfrac{BC}{CP}$ .

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Mathematics

In the following figure, DE || AC and DC || AP. Prove that : $\dfrac{BE}{EC} = \dfrac{BC}{CP}$ .

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Answer

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Mathematics

In the following figure, DE || AC and DC || AP. Prove that : BEEC=BCCP\dfrac{BE}{EC} = \dfrac{BC}{CP}ECBE​=CPBC​.

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Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that: $\dfrac{AB}{PQ} = \dfrac{AD}{PM}$ .

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