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
In Δ CED and ΔBET⇒ ∠ CED≅ ∠ BET (opposite angles)
⇒ ∠ CDE≅ ∠ BTE (Alternate angles)
(∵ AB||DC ⇒ BT||DC, as BT is extension to AB)
⇒ Δ CED ~ Δ BET (By AA Test)
$\Rightarrow \frac{ CE }{ DE }=\frac{ BE }{ TE }$ (corresponding sides are proportional)
⇒ DE× BE = CE×TE
⇒ ∠ CDE≅ ∠ BTE (Alternate angles)
(∵ AB||DC ⇒ BT||DC, as BT is extension to AB)
⇒ Δ CED ~ Δ BET (By AA Test)
$\Rightarrow \frac{ CE }{ DE }=\frac{ BE }{ TE }$ (corresponding sides are proportional)
⇒ DE× BE = CE×TE
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