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