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Question
The diagonal BD of a parallelogram ABCD bisects angles B and D. Prove that ABCD is a rhombus.
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Solution
Given: ABCD is a parallelogram where the diagonal BD bisects
parallelogram ABCD at angle B and D
To Prove: ABCD is a rhombus
Proof: Let us draw a parallelogram ABCD where the diagonal BD bisects the parallelogram at an angle B and D.
Construction: Let us join AC as a diagonal of the parallelogram ABCD
Since ABCD is a parallelogram in which diagonal BD bisects ∠B and ∠D
BD bisect ∠B,
∠ABD = ∠CBD = `1/2`∠ABC
BD bisect ∠D,
∠ADB = ∠CDB = `1/2`∠ABC
∴ ∠ABC = ∠ADC ...(Opposite angles of parallelogram are equal)
∴ `1/2`∠ABC = `1/2` ∠ADC
∠ABD = ∠ADB and ∠CBD = ∠CDB
∴ AD = AB and CD = BC
As ABCD is a parallelogram, opposite sides are equal
AB = CD and AD = BC
AB = BC = CD = AD
∴ ABCD is a rhombus.
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