Question

꯳ABCD 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

We know,
In parallelogram ABCD, AB || CD,
=> TA||CD
=> ∠TBE = ∠DCE ( Alternate angles)  ---(1)

In Δ TBE & Δ DCE ,
∠TBE = ∠DCE  (Using Equation 1)
∠BET =∠ CED  (Vertically Opposite angles) 
Δ TBE ~ Δ DCE  (AA)  

=>Both Triangles are similar .
=> Corresponding sides of similar triangles are proportional .

 $=\ \textgreater \ \frac{TE}{DE} = \frac{BE}{CE} \\ \\ =\ \textgreater \ DE*BE =TE*CE$ 

 Hence Proved :