I understand the following properties of the parallelogram:
- Opposite sides are parallel and equal in length.
- Opposite angles are equal.
- Adjacent angles add up to 180 degrees therefore adjacent angles are supplementary angles. (Their sum equal to 180 degrees.)
- The diagonals of a parallelogram bisect each other.
With that being said, I was wondering if within parallelogram the diagonals bisect the angles which the meet.
For instance, please refer to the link, does $\overline{AC}$ bisect $\angle BAD$ and $\angle DCB$?
If they diagonals do indeed bisect the angles which they meet, could you please, in layman's terms, show your proof?
Thanks, guys!
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$\begingroup$The diagonals bisect angles only if the sides are all of equal length.
Proof -
Assume that the diagonals indeed bisect angles.
Then $\angle BCE=\angle ECD$ in your diagram.
Also $\angle ECD=\angle EAB$ since $AB \| DC$.
So $\angle BCE = \angle EAB$, hence $\triangle BAC$ is isosceles with $AB=BC$.
Similar arguments also prove equality of other sides, $BC=CD$ and $CD=DA$.
$\endgroup$ $\begingroup$Nope it doesn't... We can prove it using congruency triangle. U can see that vertically opposite angles are equal. Not the diagonal bisect angle
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