1. BD is one of the diagonals of a quad ABCD .AM & CM are the perpendiculars from A & C,on BD. Show that ar(ABCD)=1/2BD(AM+CN).

2. ABCD is a trap. in which AB||CD.E is the midpoint of AD.If F is a point on BC such that segment EF is parallel to side DC, show that EF =1/2(AB+CD).



The Brainliest Answer!
1]see ar abcd = ar abd + ar bcd 
   ar abd=1/2*am*bd
ar bcd=1/2*cn*bd
ar abcd=1/2*am*bd+1/2cn*bd
ar abcd=1/2 bd[am+cn]

2]extend ad and bc to intersect at o  
now a na b are mid points of od and oc resp .
also ef //dc so f is mid point 
in tri ods ef = 3/4 dc

1 5 1
plzz mark as best
thank u so much....this helped me a lot....
ur second answer is incomplete
u said now mark as best
Area of ΔABD = 1/2 × BD × AM 
Area of ΔBCD = 1/2 × BD × CN

ARea of quad ABCD = Area of ΔABD + Area of ΔBCD = 
                                  = 1/2 × BD × AM + 1/2 × BD × CN  
                                  = 1/2 × BD ( AM +  CN)

2) Draw the diagonal BD such that it cuts EF at G

Consider ΔADB
EG parallel AB (Since EF || AB)
E is mid point of AD (given)
By converse of Basic Proportionality Theorem (BPT)
G is the mid point of BD   (1)

EG = 1/2 (AB)   ...........I
Consider ΔBCD
GF || CD (since AB || CD and EF || AB)
G is midpoint of BD (from (1))
By converse of BPT
F is midpoint of BC

By BPT, 
GF = 1/2 (CD) .......II

In trap ABCD,
EF = EG + GF
      = 1/2 (AB) + 1/2 (CD)
      = 1/2 (AB + CD)

Mark this as best since the answers are provided with beautiful images
sry dear...u r late...my maths xams r over...