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Asked by Latha Last Modified
Answer 
Tasneem
Let us join AC.
In ΔABC,
BC = AB (Sides of a rhombus are equal to each other)
∴ ∠1 = ∠2 (Angles opposite to equal sides of a triangle are equal)
However, ∠1 = ∠3 (Alternate interior angles for parallel lines AB and CD)
⇒ ∠2 = ∠3
Therefore, AC bisects ∠C.
Also, ∠2 = ∠4 (Alternate interior angles for || lines BC and DA)
⇒ ∠1 = ∠4
Therefore, AC bisects ∠A.
Similarly, it can be proved that BD bisects ∠B and ∠D as well.
read less Deleted User
Setting the foundation right through fundamentals.
Rhombus is a parallelogram with all sides of the same length.
Just consider diagonal AC. It is intercept of two parallel lines. Thus the alternate interior sides are equal. Hence Angle CAB and Angle ACD are equal.
Triangle ABC is an isosceles triangle. Thus Angle BAC and Angle BCA are equal.
Combining both we get Angle ACB, and Angle ACD is equal. Thus Diagonal AC bisects Angle C.
Similar are all the rest.
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