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If a diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.

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Solution.
Let ABCD is a parallelogram and diagonal AC bisect the angle A and \angle CAB = \angle CAD

To Prove: ABCD is a rhombus
Proof: Here it is given that ABCD is a parallelogram.
 AB\parallel CD and AC is transversal
\angle CAB=\angle ACD {alternate interior angles} …. (1)

Also AD\parallel BC and AC is a transversal
\angle CAD=\angle ACB {alternate interior angle} …. (2)
Now it is given that \angle CAB = \angle CAD

So from equations (1) and (2)
 \Rightarrow \angle ACD=\angle ACB    …. (3)
Also, \angle A=\angle C {opposite angles of a parallelogram}
\frac{\angle A}{2}=\frac{\angle C}{2}           {dividing by 2}

\angle DAC=\angle DCA {from 1 and 2}
CD=AD
{sides opposite of equal angles in \triangle ADCare equal}
AB=CD & AD=BC
{\because opposite sides of a parallelogram}

Hence,
AB = BC = CD = AD
Thus all sides are equal so ABCD is a rhombus.
Hence Proved.

 

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