ABCD is a quadrilateral in which AB || DC and AD = BC. Prove that ∠A = ∠B and ∠C = ∠D.
Solution:
Given, ABCD is a quadrilateral
AB || DC
AD = BC
We have to prove that ∠A = ∠B and ∠C = ∠D.
Extend AB to E and join CE such that AD || CE
So, AECD is a parallelogram.
We know that the opposite sides of a parallelogram are parallel and congruent.
AD = EC
Given, AD = BC
So, BC = EC
We know that the angles opposite to equal sides are equal.
∠CBE = ∠CEB ----------- (1)
We know that the linear pair of angles is always supplementary.
∠B + ∠CBE = 180° -------- (2)
Since AD || EC and cut by transversal AE,
We know that two parallel lines are cut by a transversal, the sum of interior angles lying on the same side of the transversal is always supplementary.
∠A + ∠CEB = 180° --------- (3)
From (1),
∠A + ∠CEB = 180° ------------- (4)
Comparing (2) and (4),
∠A = ∠B
We know that the sum of supplementary angles of a parallelogram is always equal to 180 degrees.
∠A + ∠D = 180°
∠B + ∠C = 180°
On comparing,
∠A + ∠D = ∠B + ∠C
since , ∠A = ∠B
∠A + ∠D = ∠A + ∠C
∠C = ∠D
Therefore, it is proven that ∠A = ∠B and ∠C and ∠D
✦ Try This: In a trapezium ABCD with AB parallel to CD, the diagonals intersect at P. The area of △ABP is 72 square cm and of ΔCDP is 50 square cm. Find the area of the trapezium.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 8
NCERT Exemplar Class 9 Maths Exercise 8.4 Problem 8
ABCD is a quadrilateral in which AB || DC and AD = BC. Prove that ∠A = ∠B and ∠C = ∠D.
Summary:
A parallelogram is a two-dimensional geometrical shape, whose sides are parallel to each other and equal in length. ABCD is a quadrilateral in which AB || DC and AD = BC. It is proven that ∠A = ∠B and ∠C = ∠D
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