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Theorem I: In a parallelogram, opposite sides are equal.

Given: A parallelogram $$ABCD$$ with $$AC$$ as its diagonal.

To prove: $$\Delta ABC\cong\Delta ADC$$.

Proof: We know that 'Opposite sides of a parallelogram are parallel'. So, $$AB||DC$$ and $$AD||BC$$.

Since $$AB||DC$$ & $$AC$$ is the transversal.

$$∠BAC =∠DCA$$ (Alternate angle ...1(yellow)).

Since $$AD||BC$$ & $$AC$$ is the transversal.

$$∠DAC =∠BCA$$ (Alternate angle ...2(red)).

In $$△ABC$$ & $$△ ADC$$.

$$∠BAC =∠DCA$$ (From alternate angle ...1).

$$∠DAC =∠BCA$$ (From alternate angle ...2).

$$AC=AC$$ (common in both).

Thus, by the $$ASA$$ criterion, the two triangles are congruent, which means that the corresponding sides must be equal.

Therefore, $$△ABC ≅ △ ADC$$ (ASA congruency).

Hence, $$AB=CD$$ & $$AD=BC$$ (corresponding parts of congruent triangles).

Hence, it is proved.

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

Given: A parallelogram $$ABCD$$ with $$AC$$ as its diagonal.

To prove: $$ΔABC ≅ ΔADC$$.

Proof: We know that 'Opposite sides of a parallelogram is parallel'. So, $$AB||DC$$ and $$AD||BC$$.

Since $$AB||DC$$ & $$AC$$ is the transversal.

$$∠BAC =∠DCA$$ (Alternate angle ...1(yellow)).

Since $$AD||DC$$ & $$AC$$ is the transversal.

$$∠DAC =∠BCA$$ (Alternate angle ...2(red)).

In $$△ABC$$ & $$△ ADC$$.

$$∠BAC =∠DCA$$ (From alternate angle ...1).

$$∠DAC =∠BCA$$ (From alternate angle ...2).

From the above figure, we can write as follows:

$$∠BAC +∠BCA$$ $$=$$ $$∠DAC + ∠DCA$$…(1)
Apply angle sum property to the triangle $$ABC$$.

$$∠B+∠BAC+∠BCA =180°$$…(2)

Now apply angle sum property to the triangle $$ACD$$.

$$∠D+∠DAC+∠DCA=180°$$…(3)

Comparing (2) and (3) with (1) we have:

$$∠B = ∠D$$.

From the theorem I, $$AB=CD$$ & $$AD=BC$$.

Thus, by the $$SAS$$ criterion, the two triangles $$ABC$$ and $$ADC$$ are congruent, which means that the corresponding sides must be equal.

Therefore, $$△ABC ≅ △ ADC$$.

Hence, it is proved.