Question

ABCD is a Rhombus, to prove 1.) AC bisects angle A and C
2.) BD bisects angle B and D

Answer 1

Answer:

Here is your answer

Let ABCD be a rhombus whose diagonals 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.

So, the diagonals of a rhombus bisect each other.

Therefore, OA = OC and OB = OD

From triangle COB and triangle COD, we have :

CB = CD ( sides of a rhombus )

CO = CO ( common )

OB = OD ( proved )

triangle COB ~ triangle COD ( by SSS congruence )

= angle COB = angle COD.

But, angle COB + angle COD = 2 right angles ( linear pairs )

angle COB = angle COD = 1right angle.

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

Hope this helps.

Plz mark it as BRAINLIEST