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Question
ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
- ABCD is a square
- diagonal BD bisects ∠B as well as ∠D.
Solution
(i) It is given that ABCD is a rectangle.
∴ ∠A = ∠C
⇒ `1/2∠A` = `1/2∠C`
⇒ ∠DAC = ∠DCA ...(AC bisects ∠A and ∠C)
CD = DA ...(Sides opposite to equal angles are also equal)
However, DA = BC and AB = CD ...(Opposite sides of a rectangle are equal)
∴ AB = BC = CD = DA
ABCD is a rectangle and all of its sides are equal.
Hence, ABCD is a square.
(ii) Let us join BD.
In ΔBCD,
BC = CD ...(Sides of a square are equal to each other)
∠CDB = ∠CBD ...(Angles opposite to equal sides are equal)
However, ∠CDB = ∠ABD ...(Alternate interior angles for AB || CD)
∴ ∠CBD = ∠ABD
⇒ BD bisects ∠B.
Also, ∠CBD = ∠ADB ...(Alternate interior angles for BC || AD)
⇒ ∠CDB = ∠ABD
∴ BD bisects ∠D.
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