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Question
A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.
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Solution
Given: Let ABCD is a parallelogram and diagonal AC bisects the angle A.
∴ ∠CAB = ∠CAD ...(i)
To show: ABCD is a rhombus.
Proof: Since, ABCD is a parallelogram, therefore AB || CD and AC is a transversal.
∴ ∠CAB = ∠ACD ...[Alternate interior angles]
So, ∠ACD = ∠ACB [∵ ∠CAB = ∠CAD, given] ...(ii)
Also, ∠A = ∠C ...[Opposite angles of parallelogram are equal]
⇒ `1/2`∠A = `1/2`∠C ...[Dividing both sides by 2]
⇒ ∠DAC = ∠DCA ...[From equations (i) and (ii)]
⇒ CD = AD ...[Sides opposite to the equal angles are equal]
But AB = CD and AD = BC ...[Opposite sides of parallelogram are equal]
∴ AB = BC = CD = AD
Thus, all sides are equal.
So, ABCD is a rhombus.
Hence proved.
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