I understand the following properties of the parallelogram:

  1. Opposite sides are parallel and equal in length.
  2. Opposite angles are equal.
  3. Adjacent angles add up to 180 degrees therefore adjacent angles are supplementary angles. (Their sum equal to 180 degrees.)
  4. 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.

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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!

  • $\begingroup$ Yes but not a normal parallelogram, only in a square, rhombus and kite. $\endgroup$ – A---B Jun 3 '16 at 17:11

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 $CD=DA$.


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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