Question

prove that diagonal of a rectangle or parallelogram divides it into two congruent triangles​

Answer 1

Answer:

In ΔBAC and ΔDCA

1 = 2

3 = 4 [alternate pair]

AC = AC (common)

Therefore, diagonal of a parallelogram divides it into two congruent triangles.

Answer 2

Step-by-step explanation:

Given :

A Parallelogram ABCD and AC is its Diagonal

Prove :

∆ ABC = ∆ CDA

Proof :

In ∆ ABC and ∆ CDA

< DAC = < BCA ( alternate Interior Angle )

So AD || BC

AC = AC ( Common side )

< BAC = < DAC ( alternate Interior Angle )

So AD || BC

< BAC = < DAC ( alternate Interior Angle )

So AB || DC

Therefore By ASA Congruence axiom

So ∆ ABC Congruent ∆ CDA