NCERT Solutions
Class 9 Maths
Quadrilaterals

Ex.8.1 Q.12
ABCD is a trapezium in which AB || CD and AD = BC (see the given figure). Show that:
(1) ∠A = ∠B
(2) ∠C = ∠D
(3) ∆ABC ≅ ∆BAD
(4) diagonal AC = diagonal BD
[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point
E. It is clear that AECD is a parallelogram.
(1) AD = CE
(Opposite sides of parallelogram AECD)
However, AD = BC
(Given)
Therefore, BC = CE
Hence, ∠CEB = ∠CBE
(Angle opposite to equal sides are also equal)
Consider parallel lines AD and CE.
AE is the transversal line for them.
∠A + ∠CEB = 1800
(Angles on the same side of transversal)
∠A + ∠CBE = 1800
(Using the relation ∠CEB = ∠CBE) .............(1)
However, ∠B + ∠CBE = 1800
(Linear pair angles) ................(2)
From equations (1) and (2), we obtain
∠A = ∠B
(2) AB || CD
∠A + ∠D = 1800
(Angles on the same side of the transversal)
Also, ∠C + ∠B = 1800
(Angles on the same side of the transversal)
=> ∠A + ∠D = ∠C + ∠B
However, ∠A = ∠B
[Using the result obtained in (1)]
=> ∠C = ∠D
(3) In ∆ABC and ∆BAD,
AB = BA
(Common side)
BC = AD
(Given)
∠B = ∠A
(Proved before)
So, ∆ABC ≅ ∆BAD
(SAS congruence rule)
(4) We have observed that,
∆ABC ≅ ∆BAD.
So, AC = BD
(By CPCT)