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Prove that parallelograms on the same base and between the some parallels are equal is area.​

Answer 1

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step-by-step explanation:

Now, in two triangles, △ ADF and △ BEC: 

∠ 1 = ∠ 4 (Corresponding angles,

AD // BC and FC is a transveral...... (Statement 1) 

AF = BE (Opposite sides of parallelogram ABEF are equal)............ (Statement 2) 

∠ 3 = ∠ 2 (Corresponding angles,

AF // BE and FC is a transversal)........ (Statement 3) 

From statement 1 and statement 3, we get: 

∠ 5 = ∠ 6 (Angle sum property) ,.......(statement 4) 

From statement 2, 3 and 4;

By ASA rule of congruency,

its' proved that: 

△ ADF ≅ △ BEC 

Since,

we know that,

areas of congruent figures are equal;

we get: 

Area of △ ADF = Area of △ BEC 

Adding area of quadrilateral ABED on both sides and we get: 

Area of △ ADF + Area of quadrilateral ABED = Area of △ BEC + Area of quadrilateral ABED 

Area of △ ADF + Area of quadrilateral ABED = Parallelogram ABEF 

Area of △ BEC + Area of quadrilateral ABED = Parallelogram ABCD

 So apply the values and we get: 

Parallelogram ABEF = Parallelogram ABCD 

Hence,

this proves the property parallelograms on the same base and between same parallel lines are equal in areas.

Answer 2

Step-by-step explanation:

Now, in two triangles, △ ADF and △ BEC:

∠ 1 = ∠ 4 (Corresponding angles,

AD // BC and FC is a transveral...... (Statement 1)

AF = BE (Opposite sides of parallelogram ABEF are equal)............ (Statement 2)

∠ 3 = ∠ 2 (Corresponding angles,

AF // BE and FC is a transversal)........ (Statement 3)

From statement 1 and statement 3, we get:

∠ 5 = ∠ 6 (Angle sum property) ,.......(statement 4)

From statement 2, 3 and 4;

By ASA rule of congruency,

its' proved that:

△ ADF ≅ △ BEC

Since,

we know that,

areas of congruent figures are equal;

we get:

Area of △ ADF = Area of △ BEC

Adding area of quadrilateral ABED on both sides and we get:

Area of △ ADF + Area of quadrilateral ABED = Area of △ BEC + Area of quadrilateral ABED

Area of △ ADF + Area of quadrilateral ABED = Parallelogram ABEF

Area of △ BEC + Area of quadrilateral ABED = Parallelogram ABCD

So apply the values and we get:

Parallelogram ABEF = Parallelogram ABCD

Hence,

this proves the property parallelograms on the same base and between same parallel lines are equal in areas.