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April 11, 2024**Theorems on Parallelograms: **If we put the sharp tip of a pencil on a sheet of paper and move from one point to the other without lifting the pencil, then the shapes so formed are called plane curves. A curve that does not cross itself at any point is called a simple curve. A simple closed plane curve made up entirely of line segments is called a polygon.

One such polygon is a parallelogram. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. In this article, we will learn in detail about the theorems of parallelograms and the properties of parallelogram.

A parallelogram is a quadrilateral with two pairs of parallel sides.

**Some Properties of a Parallelogram**

- The opposite sides and angles are equal.
- The adjacent angles are supplementary.
- The diagonals of a parallelogram bisect each other.
- Each diagonal separates it into two congruent triangles.

**Theorem**1: **A diagonal of a parallelogram divides it into two congruent triangles.**

**Given: **A parallelogram \(ABCD\) and diagonal \(AC\) divide it into two triangles \(△ABC\) and \(△CDA.\)

**To prove: **\(△ABC≅△CDA\)

**Proof: **

**Theorem **2:** In a parallelogram, opposite sides are equal.**

**Given: **A parallelogram \(ABCD.\)

**To prove: **\(AB=CD\) and \(BC=AD\)

**Construction: **Join \(AC\)

**Proof:**

**Theorem **3:** In a parallelogram, opposite angles are equal.**

**Given: **A parallelogram \(ABCD.\)

**To prove: **\(∠A=∠C\) and \(∠B=∠D\)

**Proof:**

Statements | Reasons |

\(\angle A + \angle B = {180^{\rm{o}}}\) | \(AD\parallel BC\) and \(AB\) is transversal, and the sum of co-interior angles are supplumentary |

\(\angle B + \angle C = {180^{\rm{o}}}\) | \(AB\parallel DC\) and \(BC\) is transversal, and the sum of co-interior angles are supplementary |

\(\angle A + \angle B = \angle B + \angle C\) \( \Rightarrow \angle A = \angle C\) Similarly, \(\angle B = \angle D\) | From the above two statements |

**Theorem **4:** The diagonals of a parallelogram bisect each other.**

**Given: **A parallelogram \(ABCD\) whose diagonal \(AC\) and \(BD\) intersect at \(O.\)

**To prove:** \(OA=OC\) and \(OB=OD\)

**Proof:**

NOTE: The theorems on parallelograms answer key is available on the Embibe website and app for the use of candidates.

**Theorem:** If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.

**Given: **A \(△ABC\) and a parallelogram \(BCDE\) are on the same base \(BC\) and between the same parallel lines \(BC\) and \(AD.\)

**To prove: **Area of \(\Delta ABC = \frac{1}{2}\) area of \(∥ gm\) \(BCDE\)

**Construction: **Draw \(AL⊥BC,\) meeting \(BC\) in \(L,\) and \(AL⊥BC,\) meeting \(BC\) produced in \(M.\)

**Parallelogram Theorem Proof: **Since, \(BC∥AD,\) we have \(AL=DM\)

Therefore, the area of \(\Delta ABC = \frac{1}{2} \times BC \times AL = \frac{1}{2} \times BC \times DM\)

\( = \frac{1}{2}\) area of \(∥ gm\) \(BCDE\)

Hence, the area of \(\Delta ABC = \frac{1}{2}\) area of \(∥ gm\) \(BCDE.\)

**Theorem:** Triangles on the same base and between the same parallel lines are equal in area.

**Given: **Two triangles \(△ABC\) and \(△DBC\) are on the same base \(BC\) and between the same parallel lines \(BC\) and \(FE.\)

**To prove:** Area of \(△ABC=\)Area of \(△DBC\)

**Construction: **Draw \(CE∥BA\) and \(BF∥CD\)

**Proof:**

Statements | Reasons |

\(ABCE\) and \(FBCD\) are parallelograms | By construction |

Therefore, \({\rm{Area}}\,{\rm{of}}\,\parallel {\rm{gm}}\,ABCE = {\rm{Area}}\,{\rm{of}}\,\parallel {\rm{gm}}\,FBCD\) | Parallelogram between the same base and between the same parallel lines are equal in area |

Area of \(\Delta ABC = \)Area of \(\Delta ACE \) | \(AC\) is the diagonal of the parallelogram, and a diagonal bisects the parallelogram into two congruent triangles |

Area of \(\Delta ABC = \frac{1}{2}\)Area of \(\parallel {\rm{gm}}\,ABCE\) | The area of a triangle is half that os a parallelogram on the same base and between the same parallel lines |

Similarly, the Area of \(\Delta DBC = \frac{1}{2}\) Area of \(\parallel {\rm{gm}}\,FBCD\) | The area of triangle is half that of a parallelogram on the same base and between the same parallel lines |

Hence, Area of \(\Delta ABC = \)Area of \(\Delta ACD\) | From statements \(2,\,4\) and \(5\) |

**Theorem: A diagonal of a parallelogram divides it into two triangles of equal areas.**

**Given: **\(ABCD\) is a parallelogram, and \(AC\) is one of its diagonal

**To Prove: **Area of \(△ABC=\)Area of \(△ACD\)

**Proof:**

** Q.1. In the below-given figure, **\(ABCD\)

** Ans:** Given, \(ABCD\) is a parallelogram, and opposite sides of a parallelogram are equal.

Therefore,

\(3x-1=2x+2\)

\( \Rightarrow x = 3\)

Also, opposite angles are equal,

\(\angle D = \angle B = {102^{\rm{o}}}\)

Now, for \(\Delta ACD,y = {50^{\rm{o}}} + \angle D\) (Exterior angle equals the sum of interior opposite angles.)

\( \Rightarrow y = {50^{\rm{o}}} + {102^{\rm{o}}} = {152^{\rm{o}}}\)

\(\angle DAB + {102^{\rm{o}}} = {180^{\rm{o}}}\) (\(AD∥BC,\) the sum of co-interior angles are supplementary)

\(\angle DAB = {180^{\rm{o}}} – {102^{\rm{o}}} = {78^{\rm{o}}}\)

From the figure, \(z = \angle DAB – \angle DAC = {78^{\rm{o}}} – {50^{\rm{o}}} = {28^{\rm{o}}}.\)

** Q.2. Two opposite angles of a parallelogram are **\({(3x – 2)^{\rm{o}}}\)

\( \Rightarrow {(3x – 2)^{\rm{o}}} = {(63 – 2x)^{\rm{o}}}\)

\( \Rightarrow 3x – 2 = 63 – 2x\)

\( \Rightarrow 5x = 65\)

\( \Rightarrow x = 13\)

One angle of the parallelogram\( = {(3 \times 13 – 2)^{\rm{o}}} = {37^{\rm{o}}}\)

Then the adjacent angle of the parallelogram\(1 = {180^{\rm{o}}} – {37^{\rm{o}}} = {143^{\rm{o}}}\)

Hence, all the angles of the parallelogram are \({37^{\rm{o}}},{143^{\rm{o}}},{37^{\rm{o}}},\) and \({143^{\rm{o}}}.\)

** Q.3. In the adjoining, **\(ABCD\)

** Ans:** Given, \(ABCD\) is a parallelogram,

\(3x-4=y+5\) (Opposite sides of a parallelogram are equal)

\( \Rightarrow 3x – y – 9 = 0 \ldots {\rm{(i)}}\)

And, \(2x+5=y-1\) (Opposite sides of a parallelogram are equal)

\( \Rightarrow 2x – y + 6 = 0 \ldots {\rm{(ii)}}\)

On subtracting \({\rm{(ii)}}\) from \({\rm{(i)}},\) we get,

\(x – 15 = 0 \Rightarrow x = 15\)

On substituting this value of x in the equation (i), we get,

\(3 \times 15 – y – 9 = 0 \Rightarrow 36 – y = 0 \Rightarrow y = 36\)

Therefore, \(AB = 3x – 4 = 3 \times 15 – 4 = 41\)

And, \(BC = y – 1 = 36 – 1 = 35\)

Hence, \(AB:BC = 41:35\)

** Q.4. The angles between two altitudes of a parallelogram through the vertex of an obtuse angle of a parallelogram is** \({\rm{6}}{{\rm{0}}^{\rm{o}}}.\)

** Ans:** \(ABCD\) is a parallelogram, such that \(DM⊥AB, DN⊥BC\) and \(\angle MDN = {60^{\rm{o}}}\)

Now, in quadrilateral \(DMBN,\)

\(\angle MDN + \angle M + \angle N + \angle B = {360^{\rm{o}}}\)

\( \Rightarrow {60^{\rm{o}}} + {90^{\rm{o}}} + {90^{\rm{o}}} + \angle B = {360^{\rm{o}}}\)

\( \Rightarrow \angle B = {120^{\rm{o}}}\)

Now, \(AD∥BC\) and \(AB\) is a transversal.

Therefore, \(\angle A + \angle B = {180^{\rm{o}}}\)

\( \Rightarrow \angle A + {120^{\rm{o}}} = {180^{\rm{o}}}\)

\( \Rightarrow \angle A = {180^{\rm{o}}} – {120^{\rm{o}}}\)

\( \Rightarrow \angle A = {60^{\rm{o}}}\)

We know that the opposite angles of a parallelogram are equal,

Thus, \(∠C=∠A\) and \(∠D=∠B\)

\( \Rightarrow \angle C = {60^{\rm{o}}}\) and \(\angle D = {120^{\rm{o}}}\)

Hence, the angles of the parallelogram \(ABCD\) are \({60^{\rm{o}}},{120^{\rm{o}}},{60^{\rm{o}}},\) and \({120^{\rm{o}}}.\)

** Q.5. In parallelogram **\(ABCD,AB = 10\;{\rm{cm,}}\)

As \(AE\) is the bisector of \(∠A,\)

\(\angle 1 = \angle 2 \ldots ({\rm{i}})\)

Since, \(ABCD\) is a parallelogram, \(AD∥BC,\) i.e., \(AD∥BF\)

\(∠1=∠3\) (Alternate angles are equal)

\(\angle 2 = \angle 3({\rm{Using(i)}})\)

In \(△ABF, ∠2=∠3\)

\( \Rightarrow BF = AB\)

We know that side opposite to equal angles are also equal,

Therefore,

\(BC + CF = 10\;{\rm{cm}}\)

\(AD + CF = 10\;{\rm{cm}}\,(BC = AD\) opposite sides of a parallelogram are equal)

\(6\;{\rm{cm}} + {\rm{CF}} = 10\;{\rm{cm}}\)

\( \Rightarrow {\rm{CF}} = 10 – 6\;{\rm{cm}} = 4\;{\rm{cm}}\)

Hence, the length \(CF\) is equal to \(4\;{\rm{cm}}.\)

In this article, we first had a quick view of the definition of a parallelogram: a polygon of four sides. Some properties of a parallelogram are – they have two pairs of parallel lines, opposite angles equal and adjacent angles supplementary. Then we proceeded further with the theorems on a parallelogram and learnt some theorems on parallelograms. Later we learnt the theorem based on a triangle and a parallelogram when they lie on the same parallel lines. Lastly, we solved some examples by using the theorems to strengthen our grip over the theorems on parallelograms calculator.

**Q.1: What are the theorems on different parallelograms?****Ans:** The theorems on different parallelograms are stated below.

1. A diagonal of a parallelogram divides it into two congruent triangles.

2. In a parallelogram, opposite sides are equal.

3. In a parallelogram, opposite angles are equal.

4. The diagonals of a parallelogram bisect each other.

**Q.2: How do you prove theorems on parallelograms?Ans:** We can prove this with the help of the below-mentioned theorems.

1. If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

2. If each pair of opposite angles of a quadrilateral are equal, then it is a parallelogram.

3. If a pair of opposite sides of a quadrilateral are equal and parallel, it is a parallelogram.

4. If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

**Q.3: What are the theorems on a rectangle?****Ans:** The theorem on a rectangle is mentioned below.

The diagonals of a rectangle are equal.

All the interior angles of a rectangle are right angles.

**Q.4: Are the diagonals of a parallelogram equal?****Ans:** Yes, the diagonals of a parallelogram are equal. The opposite sides and opposite angles of a parallelogram are equal. And these opposite sides and angles form two congruent triangles, with the two diagonals being the sides of these two congruent triangles. Hence, the diagonals of the parallelogram are equal.

**Q.5: What does it mean when we say that two figures are said to be on the same base and between the same parallels?****Ans:** Two figures are said to be on the same base and between the same parallel lines if they have a common base or side, and the vertices or the vertex, opposite to the common base of each figure, lie on a line parallel to the base.

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