Question 8 (ii)
ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
Diagonal BD bisects ∠B as well as ∠D.
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Solution
Let us join BD.
In Δ BCD,
BC=CD (sides of a square are equal to each other) ∠CDB=∠CBD(Angles apposite to equal sides are equal)
However, ∠CDB=∠ABD( Alternate interior angles for AB || CD) ∠CBD=∠ABD ⇒BDbisects∠B.
Also, ∠CBD=∠ADB( Alternate interior angles for BC || AD) ∠CDB=∠ABD ⇒ BD bisects ∠D