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Diagonal of a Parallelogram Which Bisects One Angle It Bisects The Second Angles

Quadrilaterals I - Concepts

Types of Quadrilateral

Angle Sum Property of Quadrilateral

Diagonal of a Parallelogram Divides it into Two Congruent Triangles

In A Parallelogram, Opposite Sides Are Equal

Opposite Angles of a Parallelogram are Equal

The Diagonals of a Parallelogram Bisect Each Other

The Bisectors of any Two Consecutive Angles of a Parallelogram Intersect at Right Angle

Diagonal of a Parallelogram Which Bisects One Angle It Bisects The Second Angles

The Angle Bisectors of a Parallelogram Form a Rectangle

Quadrilateral With One Opposite Pair Of Side Equal and Parallel is a Parallelogram

Properties of Parallelogram

Class - 9th IMO Subjects

Concept Explanation

Diagonal of a Parallelogram Which Bisects One Angle It Bisects The Second Angles

Diagonal of a Parallelogram Which Bisects One Angle Bisects The other Angle: We know that a parallelogram is a quadrilateral in which pair of opposite side is equal and parallel, but its diagonal bisects the angles of the parallelogram then that parallelogram is a rhombus.

Theorem: If diagonal of a parallelogram bisects one of the angles of the parallelogram, it also bisects the second angles. Also, prove that it is a rhombus.

GIVEN A parallelogram ABCD in which diagonal AC bisects <A.

To prove AC bisects <C

Proof Since ABCD is a parallelogram. Therefore, AB $large parallel$ DC

Now, AB $large parallel$ DC and BD intersects them,

$large therefore$ <2 = <4 .....(i) [Alternate interior angles]

Again, AD $large parallel$ BC and BD intersects them

$large therefore$ <3 = <6 ....(ii) [Alternate interior angles]

But, it is given that BD is the bisector of <B. Therefore,

<2 = <6 ....(iii)

From (i), (ii) and (iii) , we get

<3 = <4 .....(iv)Hence, BD bisects <D.

In a parallelogram <B = <D [Opposite angles are equal]

$therefore frac{1}{2}angle B = frac{1}{2}angle D$

< 2 = <3

$large Rightarrow$ AD = AB [ $large because$ Angles opposite to equal sides are equal]

But, AB = DC anf BC = AD [ $large because$ ABCD is a parallelogram]

$large therefore$ AB = BC = CD = DA

Hence, ABCD is a rhombus.

ILLUSTRATION: ABCD is a rhombus with $angle ABC= 56^0$ , then find the value of $angle ACD$

Solution: ABCD is a rhombus and each rhombus is a parallelogram

AB || CD and BC is the transversal

$therefore ;; angle ABC +angle BCD= 180^0$ ....... [ Cointerior angles are supplementary]

$Rightarrow ;; 56^0 +angle BCD= 180^0$ ......[ Substituting the value of $angle ABC$

$Rightarrow ;; angle BCD= 180^0-56^0=124^0$

Now we know that the diagonal bisects both the angles of a rhombus

$angle BCA = angle ACD$ ................[2]

From the figure

$angle BCA +angle ACD = angle BCD$

$angle ACD +angle ACD = 126^0$

$2times angle ACD = 126^0$

$angle ACD = frac{126^0}{2}= 63^0$

Sample Questions

(More Questions for each concept available in Login)

Question : 1

ABCD is a rhombus with $angle ABC= 60^0$ , then find the value of $angle ADC$ .
A. $30^{0}$	B. $60^{0}$	C. $45^{0}$	D. $90^{0}$
Right Option : B
View Explanation

Question : 2

In the following figure, ABCD is a parallelogram. The diagonal BD bisects $angle D$ . Then it must bisects ______ .
A. $angle 1$	B. $angle 2$	C. $angle B$	D. none of these
Right Option : C
View Explanation

Question : 3

In the following figure, $angle 3 =angle 4;and;angle 2=angle 6$ . Then ABCD will be a _______ .
A. Square	B. Parallelogram	C. Rhombus	D. Rectangle
Right Option : C
View Explanation