Menelaus problem

[jhonmethos]

[MATH] ABCD [/MATH] paralelogram[MATH] AE [/MATH] is the angle bisector such than [MATH]E \in DC. AE \cap BD =\{Q\} [/MATH]and [MATH]AE \cap BC =\{T\}[/MATH]. Let be a [MATH] P [/MATH] point on [MATH] (QT) [/MATH]. [MATH]DP \cap BC=\{N\}[/MATH] and [MATH] BP \cap DC=\{M\} [/MATH]. Proove that [MATH] DM=BN [/MATH]

 

[Steven G]

Please read the posting guidelines so you know how to make an appropriate post. If you had done so then you would have received help by now.

Just understand that there is a difference between a free math help forum and a math homework service website.

 

[pka]

[MATH] ABCD [/MATH] parallelogram[MATH] AE [/MATH] is the angle bisector such than [MATH]E \in DC. AE \cap BD =\{Q\} [/MATH]and [MATH]AE \cap BC =\{T\}[/MATH]. Let be a [MATH] P [/MATH] point on [MATH] (QT) [/MATH]. [MATH]DP \cap BC=\{N\}[/MATH] and [MATH] BP \cap DC=\{M\} [/MATH]. Proove that [MATH] DM=BN [/MATH]

This is a flawed question. The opposite angles of a parallelogram are congruent & diagonals are the angle bisectors. That means \(E=C\)

 

[lookagain]

The opposite angles of a parallelogram are congruent & diagonals are the angle bisectors.

The opposite angles of a parallelogram are congruent to each other. If the
parallelogram is of a special type such as a square, or even a rhombus, then
its diagonals are the angle bisectors.