In the given figure- D E A C and D F A E- Prove that ECBE - FEBF-

In ΔBAE, DF∥AE ∴ nbsp; ADDB=EFFB ............(i) nbsp; nbsp; [by Thale's theorem] In ΔBAC, DE∥AC ∴ADDB=CEEB ..............(ii) nbsp; nbsp; nbsp; n ...

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In the given ...

Question

In the given figure, DE ∥ AC and DF ∥ AE.
Prove that $\frac{E C}{B E} = \frac{F E}{B F} .$

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Solution

In ΔBAE, $DF ∥ AE$
∴ $\frac{A D}{D B} = \frac{E F}{F B}$ ............(i) [by Thale's theorem]
In ΔBAC, $D E ∥ A C$
∴ $\frac{A D}{D B} = \frac{C E}{E B}$ ..............(ii) [by Thale's theorem]
From (i) and (ii), we get:
$\frac{EF}{FB} = \frac{CE}{EB}$ $[Each equal to \frac{AD}{DB}]$
$\Rightarrow \frac{EC}{BE} = \frac{FE}{BF}$

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In the given figure, DE ∥ AC and DF ∥ AE. Prove that ECBE=FEBF.

In ΔBAE, DF∥AE ∴ ADDB=EFFB ............(i) [by Thale's theorem] In ΔBAC, DE∥AC ∴ADDB=CEEB ..............(ii) [by Thale's theorem] From (i) and (ii), we get: EFFB=CEEB Each equal toADDB ⇒ECBE=FEBF

In the given figure, DE ∥ AC and DF ∥ AE.
Prove that $\frac{E C}{B E} = \frac{F E}{B F} .$