Question

in the given figure a b parallel d e and BD parallel prove that DC square is equal to C F ×AC

Answer 1

see the upload image below

Answer 2

DC² = CF × AC

Step-by-step explanation:

The image is attached below.

If a line is parallel to one side of a triangle and the line intersects the other two sides, then this line divides those two sides proportionally.

In ΔCAB, since DE || AB,

$\frac{DC}{D A}=\frac{C E}{E B}$ (By the above theorem) ______ (1)

In ΔCDB, since EF || BD

$\frac{C F}{F D}=\frac{C E}{E B}$ (By the above theorem) ______ (2)

From (1) and (2), we get

$\frac{DC}{DA}=\frac{CF}{FD}$

Taking reciprocal on both sides,

$\Rightarrow \frac{\mathrm{DA}}{\mathrm{DC}}=\frac{\mathrm{FD}}{\mathrm{CF}}$

Add 1 on both sides.

$\Rightarrow \frac{\mathrm{DA}}{\mathrm{DC}}+1=\frac{\mathrm{FD}}{\mathrm{CF}}+1$

$\Rightarrow \frac{A D+DC}{DC}=\frac{F D+C F}{C F}$

$\Rightarrow \frac{A C}{DC}=\frac{DC}{C F}$

Do cross multiplication.

${DC}^{2}={AC} \times {CF}$

Hence proved.

To learn more...

1. In the given figure, DE || AB and FE || DB. Prove that DC2 = CF . AC

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2. In the given figure AD and BE are the medians of triangle ABC and BE IS Parallel toDF .prove that CF=1/4AC

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