Question

Prove that the line segment joining the midpoints if two non-parallel sides of a trapezium is parallel to the parallel sides and equal to half of the sum of parallel sides. ​

Answer 1

Answer:

Step-by-step explanation:

Let say ABCD is Trapezium

M & N are mid points of AD & BC

=> MN ║ AB ║ CD

Join AC which intersect MN at P

=> MP ║ AB ║ CD & NP ║ AB ║ CD as P is point on MN

Now in ΔABC

PN ║ AB

=> PN/AB = CN/BC

CN = BC/2 ( as N is mid point of BC)

=> CN/BC = 1/2

=> PN/AB = 1/2

=> PN = AB/2

Similarly in ΔACD

MP ║ CD

=> AM/AD = MP/CD

AM = AD/2 ( as M is mid point of AD)

=> AM/AD = 1/2

=> MP/CD = 1/2

=> MP = CD/2

MP + PN = AB/2 + CD/2

=> MN = (AB + CD)/2

Hence Proved that line segment joining the midpoint of two non parallel side of a Trapezium is equal to half of their sum