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 Please do not Add Links with your comments as they will not be published.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