Question

In a parallelogram ABCD, from vertex D, a line is draw which intersects produced BA and BC at E and F respectively. prove $\frac{AD}{AE}=\frac{EB}{BE}=\frac{FC}{CD}$

Answer 1

In $△EAD$ and $△DCF$

$\angle 1=\angle 2[AB\parallel CD]$

$\angle 3=\angle 4[AD\parallel BC]$

$\therefore △EAD\sim △DCF$

[By AA similarity]

$\frac{EA}{DC}=\frac{AD}{CF}=\frac{DE}{FD}$ [BySSS]

$\frac{AD}{AE}=\frac{CF}{CD}\to \left(2\right)$

In $△EAD$ and $△EBF$

$\angle 1=\angle 1$ (common

$\angle 3=\angle 4[AD\parallel BC]$

$\therefore △EAD\sim △EBF$ [By AA similarity]

$\frac{EA}{EB}=\frac{AD}{BF}=\frac{DE}{EF}$ [By SSS]

$\frac{AD}{AE}=\frac{FB}{BE}\to \left(2\right)$

From (1) and (2) , $\frac{AD}{AE}=\frac{FB}{BE}=\frac{CF}{CD}$