Question

◻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

Given: ◻ABCD is a parallelogram
To prove: DE × BE = CE × TE
Proof: In ∆BET and ∆CED
∠BET = ∠CED (Vertically opposite angles)
∠BTE = ∠CDE (Alternate angles, AT || CD and DT is a transversal line)
By AA test of similarity
∆BET ∼ ∆CED
$$\begin{gathered} \therefore \frac{\mathrm{BE}}{\mathrm{CE}}=\frac{{\displaystyle \mathrm{ET}}}{{\displaystyle \mathrm{ED}}}\left(\mathrm{Corresponding}\mathrm{sides}\mathrm{are}\mathrm{proportional}\right) \\ \Rightarrow \mathrm{BE}\times \mathrm{ED}=\mathrm{CE}\times \mathrm{ET} \end{gathered}$$
Hence proved.