In the given trapezium PQRS , A and B are the mid - points of the diagonals QS and PR respectively. Prove that : 1. AB SR

2. AB = ( SR - PQ )

[ Hint : Join Q and B and produce QB to meet SR at C ].

~Thanks in advance !

Step-by-step explanation:jj

Given: In trapezium PQRS , A and B are midpoints of diagonals PR and QS.

To Show: AB=1/2(PQ-SR).

Construction:draw midpoints C and D of SP and RQ respectively and join CA and BD

Proof : In triangle SPQ ,C and B are midpoints of SP and SQ

Therefore,by midpoint theorem

CB || PQ and 2CB= PQ 1

In triangle QSR B and D are midpoints of QS and RQ

Therefore, by midpoint theorem

BD || SR and 2BD =SR2

In triangle SPR C and A are midpoints of SP and RP

Therefore by midpoint theorem

CA = 1/2 SR 3

From 2

2 BD = SR

BD =1/2 SR 4

From 3 and 4

CA = BD 5

From 1 and 2

PQ - SR = 2 CB - 2 BD

PQ - SR = 2(CB - BD)

1/2(PQ - SR) =CB - BD

1/2(PQ - SR)= CA+AB - BD

From 5

CA = BD , therefore they cancel each other

Therefore ,

1/2(PQ-SR) =AB

AB= 1/2(PQ-SR)

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

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