Parallelogram theorem-I, II — lesson. Mathematics State Board, Class 9.

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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)).

$△ABC$ &

$△ ADC$ .

$∠BAC =∠DCA$ (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)).

$△ABC$ &

$△ ADC$ .

$∠BAC =∠DCA$ (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.

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3. Parallelogram theorem-I, II

Theory: