Question

Prove that the median of a trapezium is parallel to its parallel sides and is equal to one half of their sum.

Answer 1

Answer:

Step-by-step explanation:ABCD is a trapezium in which AB || DC and E, F are mid points of AD, BC respectively. Join CE and produce it to meet BA produced at G. In ΔEDC and ΔEAG, ED = EA ( E is mid point of AD) ∠CED = ∠GEC ( Vertically opposite angles) ∠ECD = ∠EGA ( alternate angles) ( DC||AB, DC||GB and CG transversal) ∴ ΔEDC ≅ ΔEAG CD = GA and EC = EG In ΔCGB, E is mid point of CG ( EC = EG proved) F is a mid point of BC (given) ∴ By mid point theorem EF ||AB and EF = (1/2)GB. But GB = GA + AB = CD + AB Hence EF||AB and EF = (1/2)( AB + CD).