A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.
Let the parallelogram be = ABCD
Diagonal AC bisect ∠A.
∠CAB = ∠CAD
Now,
AB||CD and AC is a transversal.
∠CAB = ∠ACD
Again, AD||BC and AC is a transversal.
∠DAC = ∠ACB
Now,
∠A = ∠C
½ ∠A = ½ ∠C
∠DAC = ∠DCA
AD = CD
But, AB = CD and AD = BC (Opposite sides of parallelograms)
AB = BC = CD = AD
Thus, ABCD is a rhombus.
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