Question

prove that the diagonals of a Rhombus bisect each other at right angle

Answer 1

For solving it you have to first prove this two triangles supplementary and then you get the answer . look at pic you will be more clearer

Thank you

Answer 2

Given: A rhombus ABCD with diagonals AC and BD intersecting at O.

To Prove: OA = OC ; OB = OD ; ∠AOB = ∠COB = 90°

Proof: In △'s OAB and ODC

∠OAB = ∠OCD [AB ll CD ,Alternate ∠s]

∠OBA = ∠ODC [AB ll CD,Alternate ∠s]

AB = CD [Opposite sides of a rhombus]

By ASA

△OAB ≅ △ODC

➱ OA = OC,OB = OD [c.pc.t.]

Again in △'s OAB and OBC

OA = OC [Proved]

OB = OB [Common]

AB = BC [Sides of a rhombus]

By S.S.S

△OAB ≅ △OBC

➱ ∠AOB = ∠COB [c.p.c.t] .......(i)

But, ∠AOB + ∠COB = 180° [Linear pair]

∴ ∠AOB = ∠COB = 90° [From (i) ]