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2. The diagonals of a parallelogram bisect each other. D 3 2 C

Let us take a parallelogram ABCD and draw both the diagonals, O
AC and BD in it. Let the diagonals intersect each other at O as 1 4
shown in the adjoining figure. A B

In DAOB and DDOC, we have
∠1 = ∠2 (Alternate angles as AB||CD)

∠3 = ∠4 (Alternate angles as AB||CD)
AB = CD (Opposite sides of the parallelogram are equal)

Therefore, DAOB ≅ DCOD (By ASA criterion)
Thus, OA = OC and OB = OD (By CPCT)
Hence, in a parallelogram, diagonals bisect each other.

3. Adjacent angles of a parallelogram are supplementary.
Let us take a parallelogram EFGH. We know that the opposite E F
angles of a parallelogram are equal. a b

So, we have ∠E = ∠G = ∠a
And ∠F = ∠H = ∠b a
b
By using the interior angle sum property of a quadrilateral, H G
we have

∠E + ∠G + ∠F + ∠H = 360° Think and Answer
⇒ ∠a + ∠a + ∠b + ∠b = 360° In the given parallelogram, if OE = 4 cm and HL is 5 cm

⇒ 2∠a + 2∠b = 360° more than PE, then what will be the length of OH?
P L
⇒ 2(∠a + ∠b) = 360°

⇒ ∠a + ∠b = 180° O 4 cm
\ ∠E + ∠H = 180° H E

Similarly, ∠H + ∠G = 180°, ∠G + ∠F = 180°, ∠F + ∠E = 180°
Hence, in a parallelogram, the adjacent angles are supplementary, that is, their sum is 180°.
Example 18: In the adjoining figure, EFGH is a parallelogram such E F
that ∠H = 60°. Find all other angles of the parallelogram EFGH.

Solution: In a parallelogram, the opposite angles are equal, that
is, ∠H = ∠F.
60°
So, ∠F = 60° H G

Again, in a parallelogram, the adjacent angles are supplementary, that is, ∠H + ∠E = 180°.
So, 60° + ∠E = 180°
⇒ ∠E = 180° – 60° = 120°

Again, ∠E = ∠G (Opposite angles of parallelogram)
⇒ ∠G = 120°

Thus, for the given parallelogram EFGH, ∠E = 120°, ∠F = 60°, ∠G = 120°, and ∠H = 60°.

69 Quadrilaterals