Question

Prove that diagonals bisect each other at right angle in rhombus.

Answer 1

Proof . – Let ABCD be a rhombus whose diagonal AC and BD intersect at the point O. We know that the diagonals of a parallelogram bisect each other. Also we know that every rhombus is a parallelogram. Therefore OA=OC and OB=OD. From triangle(COB) and triangle (COD), we have: CB=CD sides of rhombus. CO=CO. Common OB=OD proved Therefore tri(COB)~tri(COD) by SSS congruence. =>

Answer 2

Solution :

Given -

A Rhombus ABCD whose diagonals AC and BD intersect at point O.

To prove :-

(i) OA = 0C and OB = OD

(ii) Angle BOC = Angle DOC = Angle AOD = Angle AOB = 90°

Proving :-

(i) Clearly , ABCD , is a || gm , in which AB = BC = CD = DA

Also , we know that the diagonals of a || gm bisect each other .

Hence , OA = OC and OB = OD

(ii) Now , in ∆s BOC and DOC , we have :-

OB = OD , BC = DC and OC = OC ( common )

Hence , ∆BOC ≈ ∆DOC

Hence , ∆BOC ≈ ∆DOC ( c.p.c.t. )

But , Angle BOC + Angle DOC = 180° ( linear pair )

Hence , Angle BOC = Angle DOC = 90°

Similarly , Angle AOB = Angle AOD = 90°

Hence , the diagonals of a Rhombus bisect each other at right angles.