Question

Prove this theorem by SSS rule.

"a diagonal of a parallelogram divides it into two congruent triangles"

By SSS rule only.

Answer 1

it can be proved by ASA




Given: A parallelogram ABCD and AC is its diagonal .
To prove : △ABC ≅ △CDA
Proof : In △ABC and △CDA, we have 
∠DAC =  ∠BCA [alt. int. angles, since AD | | BC] 
AC = AC [common side] 
and ∠BAC =  ∠DAC [alt. int. angles, since AB | | DC]  
∴ By ASA congruence axiom, we have 
△ABC ≅ △CDA

Answer 2

Statement : A diagonal of a parallelogram divides it into two congruent triangles.

Given : A parallelogram ABCD.

To prove : ΔBAC ≅ ΔDCA

Construction : Draw a diagonal AC.

Proof :

In ΔBAC and ΔDCA,

∠1 = ∠2 [alternate interior angles]
∠3 = ∠4 [alternate interior angles]
AC = AC [common]

ΔBAC ≅ ΔDCA [ASA]

Hence, it is proved.