Question

c ABCD is a parallelogram point E
is on side BC. Line DE intersects ray
AB in point T. Prove that
DE ´ BE = CE ´ TE.

Answer 1

I'm proving that DE x BE = CE x TE.

ABCD is a parallelogram. ∴ AD || BC and AB || CD. 
∠DCB + ∠ABC = 180°. 
If ∠DCB = x°, then ∠ABC = (180 - x)°. 
∠ABC and ∠TBC are linear pairs. 
∴ ∠TBC = (180 - (180 - x))° = (180 - 180 + x)° = x°. 
(∠DCB and ∠TBC are alternative angles also. ) 

AT is continuous of AB. ∴ AT || CD. 
∠BTD and ∠CDT are alternative angles. ∴ ∠BTD = ∠CDT. 
∠DEC = ∠BET (alternative angles). 

∴ ΔCDE ~ ΔBET. 

Sides opposite to equal angles of two similar triangles are in the same ratio. 

∴ $\frac{DE}{TE} = \frac{DC}{BT} = \frac{CE}{BE}$

On $\frac{DE}{TE} = \frac{CE}{BE}$, by cross multiplication, we get, 
DE x BE = CE x TE. 

Hope this answer will be helpful.