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 DC Now, AB DC and BD intersects them, <2 = <4 .....(i) [Alternate interior angles] Again, AD BC and BD intersects them <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] < 2 = <3 AD = AB [ Angles opposite to equal sides are equal] But, AB = DC anf BC = AD [ ABCD is a parallelogram] AB = BC = CD = DA Hence, ABCD is a rhombus. |
ILLUSTRATION: ABCD is a rhombus with , then find the value of
Solution: ABCD is a rhombus and each rhombus is a parallelogram AB || CD and BC is the transversal ....... [ Cointerior angles are supplementary] ......[ Substituting the value of Now we know that the diagonal bisects both the angles of a rhombus ................[2] From the figure |
ABCD is a rhombus with , then find the value of . | |||
Right Option : A | |||
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ABCD is a rhombus with , then find the value of . | |||
Right Option : A | |||
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In the following figure, the diagonal AC bisects . So it must bisects _____ . | |||
Right Option : B | |||
View Explanation |
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