Question

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

Answer 1


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

Step-by-step explanation:

To Prove :-

Diagonal of a parallelogram divide it into two congruent triangles .

• Solution :-

Given :-

ABCD is a parallelogram . AC is the diagonal .

To prove :-

∆ ADC congruent to ∆ ABC

Proof :-

Opposite sides of a parallelogram are parallel to each other . So,

DC || AB

AD is transverse line . So,

Angle DAC = Angle ACB .

Angle DCA = Angle CAB

In triangle ADC and triangle ABC

AD = AD. ( common )

Angle DAC = Angle ACB ( Proved above)

Angle DCA = Angle CAB ( Proved above)

By ASA criteria .

∆ ADC is congruent ∆ ABC .

Hence proved