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.
Solution:
(i) We are given that ABCD is a rectangle, so
∠A = ∠C
⇒ 1/2 ∠A = 1/2 ∠C
⇒ ∠DAC = ∠DCA (Given that AC bisects ∠A and ∠C)
Thus, CD = DA (Sides opposite to equal angles are also equal)
However, DA = BC and AB = CD (Opposite sides of a rectangle are equal)
Thus AB = BC = CD = DA
ABCD is a rectangle and all the sides are equal. Hence, ABCD is a square.
(ii) Let us join BD
Since ABCD is square, AB || CD and BC || AD
In ΔBCD,
BC = CD (Sides of a square are equal to each other)
∠CDB = ∠CBD (Angles opposite to equal sides are equal) ..... (1)
However, ∠CDB = ∠ABD (Alternate interior angles as AB || CD) ..... (2)
From equations (1) and (2), ∠CBD = ∠ABD
Thus, BD bisects ∠B.
Also, ∠CBD = ∠ADB (Alternate interior angles for BC || AD) .....(3)
So, using equations (1) and (3), ∠ADB = ∠CDB
Hence, BD bisects ∠D and ∠B.
☛ Check: Class 9 Maths NCERT Solutions Chapter 8
Video Solution:
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.
NCERT Maths Solutions Class 9 Chapter 8 Exercise 8.1 Question 8
Summary:
If ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C, we have proved that ABCD is a square and diagonal BD bisects ∠B as well as ∠D.
☛ Related Questions:
- Show that the diagonals of a square are equal and bisect each other at right angles.
- Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
- Diagonal AC of a parallelogram ABCD bisects ∠A (see the given figure). Show that i) it bisects ∠C also, ii) ABCD is a rhombus.
- ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.