Answer with explanation:
A Parallelogram ABCD, in which , AD=8 in, and AB=3 in.
Opposite sides as well aas Opposite angles of Parallelogram are equal.
∠A=∠C, ∠B=∠D,AB=CD, AD=BC
It is given that ,∠A and ∠D is bisected by two line segments which divides Side BC into three segments.
Let, ∠A= ∠1 +∠2,such that, ∠1= ∠2.
And, ∠D=∠3+∠4 , such that, ∠3=∠4.
Sum of interior angles on the same side of transversal is 180°.
∠A+∠D=180°
∠D=180°- ∠A
So,
[tex]\angle 4=90^{\circ}-\frac{\angle A}{2}\\\\\text{Using Angle sum property of triangle}\\\\\angle 4+\angle C+\angle 5=180^{\circ}\\\\90^{\circ}-\frac{\angle A}{2}+\angle A+\angle 5=180^{\circ}\\\\\angle 5=90^{\circ}-\frac{\angle A}{2}\\\\\angle 5=\angle 4[/tex]
CN=CD----[If opposite angles of triangle are equal , then side opposite to these angles are equal.]
CD=3 in
So, CN=3 in.
Similarly, AB=BM=3 cm
So, MN=8-BM-CN
=8-3-3
=2 in.
