Question

in the figure de parallel BC and CD parallel to EF prove that ad square equals to ab × af​

Answer 1

In The triangle ABC,

   DE || BC

= $\frac{AB}{AD} = \frac{AC}{AE}$   (i)

in the other triangle ADC, we have

    FE||DC

= $\frac{AD}{AF} = \frac{AC}{AE}$      (ii)

From (i) and (ii) we get

$\frac{AB}{AD} = \frac{AD}{AF}$

= $AD^{2} = AB \times AF$

Hence proved

Answer 2

Answer:

Step-by-step explanation: