□ABCD is a parallelogram, point E is on side BC. Line DE intersects ray AB in point T. Prove that DE × BE = CE × TE.
Figure: T lies on ray AB beyond B (A–B–T), and D–E–T.
Answer: \(\triangle BET \sim \triangle CED\) (AA test), which gives DE × BE = CE × TE.
Side AB ∥ side DC (opposite sides of a parallelogram), and T lies on line AB, so seg BT ∥ seg DC.
In \(\triangle BET\) and \(\triangle CED\):
\(\therefore \triangle BET \sim \triangle CED\) (AA test).
Corresponding sides are in proportion:
\[\frac{BE}{CE} = \frac{TE}{DE}\] \[\therefore DE \times BE = CE \times TE\]Hence proved.
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