Two diagonals of a rhombus bisect each other at right angles

Proposition:Let the diagonals AC and BD of the rhombus ABCD intersect at O. It is required to prove that,
(i) ∠AOB = ∠BOC = ∠COD = ∠DOA = 1 right angle
(ii) AO = CO, BO = DO.


Proof:Step-1: A rhombus is a parallelogram. Therefore, AO = CO, BO = DO. (Diagonals of a parallelogram bisect each other)

Step-2: Now in ΔAOB and ΔBOC,
AB = BC (sides of a rhombus are equal)
AO = CO
and OB = OB. (common side)
So ΔAOB = ΔBOC.
Therefore, ∠AOB = ∠BOC.
∠AOB + ∠BOC = 1 straight angle = 2 right angles.
∠AOB = ∠BOC = 1 right angle.
Similarly, it can be proved that, ∠COD = ∠DOA = 1 right angle. (Proved)
Axact

Digital STUDY Center

Digital Study Center offers an effective and amazing learning platform for keen learn students in the world. We identify the needs and demands of the keen learn students which is why we stand out unique in the crowd.

Post A Comment:

0 comments:

Dear readers,
Your feedback is always appreciated. We will reply to your queries within 24hrs. Before writing your comments, please read the following instructions attentively:

1. Please comments in English. We accept only English comments.

2. Please do not Spam. All spammed comments will be deleted as soon as pobile, after review.

3. Please do not Add Links with your comments as they will not be published.

4. If We can be of assistance, please do not hesitate to contact us.