Question

Prove that a diagonal of a parallelogram divides the parallelogram into two
congruent triangles.

Answer 1

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Diagonal of a parallelogram divide it into two congruent triangles .

${\boxed{\boxed{\mathtt{Solution}}}}$

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

Answer 2

Prove that a diagonal of a parallelogram divides the parallelogram into two  congruent triangles.

Explanation:

Given: A parallelogram ABCD with AC as its diagonal.

To prove: ΔABC ≅ ΔADC

Proof:  Since the line segments AB and CD are parallel to each other,  and AC is a traversal we have,

∠ACD = ∠CAD

AC = AC [common side]

∠CAB = ∠ACD

Thus, by angle-side- angle(ASA) congruence, we have,

ΔABC ≅ ΔACD

The corresponding parts of the congruent triangles are congruent.

Therefore, diagonal of a parallelogram divides the parallelogram into two

congruent triangles.

Hence proved.

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