Given that,
ABCD is a rhombus.
AC and BD are its diagonals.
To Prove
AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Solution:
As, ABCD is a rhombus, so
AB = BC = CD = DA
Now, As CD = AD
∠ADB= ∠ABD……………………….(1) [Angles opposite to equal sides of a triangle are equal]
∠ADB = ∠CBD …………………..(2) [ ∵ Alternate interior angles are equal] (Rhombus is a parallelogram)
From (1) and (2), we have
∠CBD = ∠ABD …………………..(3)
∠ABD= ∠CDB ……………………..(4) [ ∵ Alternate interior angles are equal]
From (1) and (4),
we have ∠ADB = ∠CDB
Hence, BD bisects ∠B as well as ∠D.
Similarly, we can prove that AC bisects ∠C as well as ∠A.
Hence Proved
