Question

In trapezium ABCD , side AB // side DC & side AD = side BC. If ∠A = 55° then find ∠B & ∠C.​

Answer 1

side BC∥ side AD

Interior angles on the same side of the transversal are supplementary.

∠A+∠B=180

∘

⇒72

∘

+∠B=180

∘

⇒∠B=180

∘

−72

∘

⇒∠B=108

∘

Construction:

Draw BP ∥⊂D

So,BC∥AD and BP∥CD

⇒PBCD is a parallelogram

⇒CD≅BP

Now CD=BP and CD=AB

⇒BP=AB

⇒∠BAP=∠BPA=72

∘

BP∥CD so,

∠CDP=∠BPA=72

∘

(corresponding angles are equal)

∠B=108

∘

∠D=72

∘

Answer 2

Answer:

Join BD, let BD and MN meet at Q. Since, M is the mid point of AD and N is the mid point of BC. So by mid point theorem, AB II MN IICD

In △, BDC and BQN,

∠B=∠B (Common)

∠BDC=∠BQN (Corresponding angles of parallel lines)

∠BCD=∠BNQ (Corresponding angles of parallel lines)

thus, △BDC∼△BQN

Thus,

QB

BD

=

QN

DC

2=

QN

DC

(Q is the mid point of BD)

QN=

2

1

DC

Similarly, QM=

2

1

AB

Hence, QM+QN=

2

1

(AB+DC)

MN=

2

1

(AB+CD)

AB+CD=2MN