Question

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

1. AB
`\parallel` SR

2. AB =
`(1)/(2)` ( SR - PQ )

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

~Thanks in advance !

Answer 1

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

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

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 II SR and 2BD =SR..2

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

SR -PQ= 2 BD -2 CB -

SR-PQ = 2( BD-CB)

From 5

CA = BD ,

Therefore,

1/2(SR-PQ) =AB

AB= 1/2(SR-PQ)

Hence Showed

Answer 2

See Below.

Explanation:

In the given trapezoid, A and B are the mid-points of QR and PR, respectively.

Question 1)

Please refer to the first diagram below.

As instructed, we will join Q and B to produce QB to meet SR at C.

Statements: Reasons:

`1)\text{ } PQ\parallel SR` Given

`2)\text{ } \angle RCQ\cong \angle PQC` Alternate Interior Angles Theorem

`3)\text{ } B \text{ is the midpoint of } PR` Given

`4)\text{ } PB=BR` Definition of Midpoint

`5)\text{ } \angle CRB\cong \angle QPB` Alternate Interior Angles Theorem

`6)\text{ } \Delta CRB\cong \Delta QPB` AAS-Congruence

`7)\text{ } QB\cong CB` CPCTC

`8)\text{ }A\text{ is the midpoint of } SQ` Given

`9)\text{ } SA=AQ` Definition of Midpoint

`10)\text{ } AB\parallel SR` Triangle Midsegment Theorem

Question 2)

(We will use the proven statements above as "given.")

(And please continue referring to the first figure provided.)

Statements: Reasons:

`1)\text{ } AB\parallel SR` Given

`2)\text{ } \angle QSR\cong\angle QAB` Corresponding Angles Theorem

`3)\text{ } \angle SCQ\cong\angle ABQ` Corresponding Angles Theorem

`4)\text{ } \Delta SCQ\sim \Delta ABQ` Angle-Angle Similarity

`5)\text{ } \displaystyle (SQ)/(AQ)=(SC)/(AB)` CSSTP

`6)\text{ } A\text{ is the midpoint of } SQ` Given

`8)\text{ } SA=AQ` Definition of Midpoint

`9)\text{ } SQ=SA+AQ` Segment Addition

`10)\text{ } SQ=2AQ` Substitution

`11)\text{ } \displaystyle (2AQ)/(AQ)=(SC)/(AB)` Substitution

`12)\text{ } \displaystyle 2=(SC)/(AB)` Division Property of Equality

`13)\text{ } 2AB=SC` Multiplication Property of Equality

`14)\text{ } SR=SC+CR` Segment Addition

`15)\text{ } \Delta CRB\cong\Delta QPB` Given

`16)\text{ } CR\cong PQ` CPCTC

`17)\text{ }SR=SC+PQ` Substitution

`18)\text{ } SC=SR-PQ` Subtraction Property of Equality

`19)\text{ } 2AB=SR-PQ` Substitution

`\displaystyle 20)\text{ } AB=(1)/(2)(SR-PQ)` Division Property of Equality