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

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Best answer

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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