Question

Prove converse of basic proportionality theorem.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Converse of Basic Proportionality Theorem:

If a line divides two sides of a triangle in the same ratio , then the line is parallel to the third side .

Given:

In ∆ABC , a line DE is drawn such that $\frac{AD}{DB}=\frac{AE}{EC}$

RTP:

$DE \parallel BC$

PROOF:

Assume that DE is not parallel to BC then draw the line DE' parallel to BC.

$So,\\\frac{AD}{DB}=\frac{AE'}{E'C}$

/* By Basic Proportionality Theorem*/

$Therefore, \\\frac{AE}{EC}=\frac{AE'}{E'C} \:[ Given ]$

Adding 1 to both sides of the above , we can see that E and E' must coinside .

$\frac{AE}{EC}+1=\frac{AE'}{E'C}+1\\=\frac{AE+EC}{EC}=\frac{AE'+E'C}{E'C}$

$\implies \frac{AC}{EC}=\frac{AC}{E'C}$

$\implies \frac{1}{EC}=\frac{1}{E'C}$

$\implies EC=E'C$

$This \: is \: possible \: only\\when \: E \: and \: E' \: coinside \\\: i.e \: DE'\: is \: the \:line \: DE \: itself.$

$Hence , DE \parallel BC$

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