$□ A B C D$ is a parallelogram point $E$ is on side $B C$ Line $D E$ intersect ray $A B$ in point $T$ .
Prove that $D E \times B E = C E \times T E$

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Solution

ABCD is a $∥ g m$ , then
$A B ∥ C D$ , therefore $A T ∥ D C$
BecauseAT is extension of AB,
In $△ T B E$ and $△ D C E$
$∠ T B E = ∠ E C D$ [Alternate angles]
$∠ B E T = ∠ C E D$ [Vertically opposite]
Therefore, $△ B E T \sim△ C E D$ [By AA similarity]
$∴ \frac{T E}{D E} = \frac{B E}{C E} = \frac{B T}{C D}$ [By CPCT]
$\Rightarrow \frac{T E}{D E} = \frac{B E}{C E} \Rightarrow T E * C E = B E * D E$

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□ABCD is a parallelogram point E is on side BC Line DE intersect ray AB in point T.Prove that DE×BE=CE×TE

ABCD is a ∥gm , thenAB∥CD, therefore AT∥DCBecauseAT is extension of AB,In △TBE and △DCE∠TBE=∠ECD [Alternate angles]∠BET=∠CED [Vertically opposite]Therefore, △BET∼△CED [By AA similarity]∴TEDE=BECE=BTCD [By CPCT]⇒TEDE=BECE⇒TE∗CE=BE∗DE

$□ A B C D$ is a parallelogram point $E$ is on side $B C$ Line $D E$ intersect ray $A B$ in point $T$ .
Prove that $D E \times B E = C E \times T E$