Question

$\huge\frak\red{Question\:❓}$


Q. The diagonal of BD of a parallelogram ABCD intersects AE at point F, where E is any point on the side BC. Prove that DF × EF = FB × FA.

$\small\cal\red{This\: question\:is\:not\ \: to\: SpaM ❌}$
​

Answer 1

Given: The diagonal BO of parallelogarn ABCD intersects the segment AE at F,

where E is any point on BC.

To provo: DF x EF= FB x FA

Proof: In triangles AFD and BFE,

∠FAD = ∠FEB (Alternate angles)

∠AFD = ∠BFE (Vertically opposite angles)

Therefore △ADF ~ △BFE (AA similarity)

DF/FA = FB/EF

Hence DF x EF = FB x FA

HERE YOURS RIGHT ANSWER....

Answer 2

kl aapka birthday hae

to happy birthday in advance dear