Question

ABCD is trapezium in which AB||DC. E is the mid point of AD. If F is point on side BC such that segment EF is parallel to side DC, show that EF = ½ (AB + CD).

Answer 1

Sol: ABCD is a trapezium in which AB ||CD and E and F is the mid point of AD and BC........!!!!! Join CE to produce it to meet BA and produced to G.... ! So now in tri.EDC and EAG,, EA = ED( E is the mid point of AD....< CED = < GEC ( ver. Opposite angles ). ...........< ECD = < EGA ( DC | | AB AND DC | | GB AND CD is a trans. ..). So therefore ..tri .EDC = tri EAG ( ASA CONGRUENCE RULE ). ..... and also CD = GA AND EC = EG( C. P.C.T ). ... In tri CGB. ,E is the mid point of AD and F IS THE MID POINT OF BC ( EC = EG prooved above ). . Therefore by mid point theorem EF | | AB AND EF = 1 / 2 OF GB. .. But GB = GA + AB = CD + AB. ...... ( GA = CD prooved above ). .... hence EF | | AB AND EF = 1 / 2 ( AB + CD ). ...

Answer 2

ABCD is trapezium in which AB||DC.
EF is parallel to side DC. Then we have AB||DC||EF
Hence we have also trapezium ABFE and trapezium EFCD
Let AP be the perpendicular to DC and this intersects EF at Q.
AQ will be perpendicular to EF.
 For triangle APD and AQE we have AD/EA=AP/AQ=2
         This gives AP=2AQ  ie AQ = QP
Consider the area  we have area ABCD= area ABFE + area EFCD                    (1/2)AP*(AB+DC)= (1/2)AQ*(AB+EF) + (1/2)QP*(EF+DC)       
 ⇒  AP(AB+DC)=  AP*AB/2 + AP*EF/2 +AP*EF/2 +AP*DC/2       
 ⇒  AP*AB/2 + AP*DC/2 = AP*EF/2 + AP*EF/2
 ⇒ AP*(AB+DC)/2 =  AP*EF
 ⇒  EF=(AB+DC)/2