Question
$J A B C D$ is a parallelogram. Point $E$ is on side $B C$. Line $D E$ intersects ray $A B$ in point $T$. Prove that $D E \times B E=C E \times T E$.
✓
Answer
$\begin{array}{l}\text { In } \triangle CED \text { and } \triangle BET \\ \Rightarrow \angle CED \cong \angle BET \text { (opposite angles) } \\ \Rightarrow \angle CDE \cong \angle BTE \text { (Alternate angles) } \\ (\because AB \| DC \Rightarrow BT \| DC \text {, as } BT \text { is extension to } AB ) \\ \Rightarrow \triangle CED \sim \triangle BET \text { (By AA Test) } \\ \Rightarrow \frac{ CE }{ DE }=\frac{ BE }{ TE } \text { (corresponding sides are proportional) } \\ \Rightarrow DE \times BE = CE \times TE \end{array}$
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