Question

ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C. Show that: 

(i) ABCD is a square 

(ii) diagonal BD bisects ∠B as well as ∠D

Answer 1

(i) It is given that ABCD is a rectangle.

∠A =∠C

\(⇒ \frac{1}{2}∠A=\frac{1}{2}∠C\)

\(⇒∠DCA=∠DCA\)   (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 ∆BCD, 

BC = CD (Sides of a square are equal to each other) 

∠CDB = ∠CBD (Angles opposite to equal sides are equal) 

However, ∠CDB = ∠BD (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.