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

2. AB = [tex] \frac{1}{2} [/tex] ( SR - PQ )

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

~Thanks in advance !

Answer:

See Below.

Step-by-step 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:

[tex]1)\text{ } PQ\parallel SR[/tex]                                                 Given

[tex]2)\text{ } \angle RCQ\cong \angle PQC[/tex]                                     Alternate Interior Angles Theorem

[tex]3)\text{ } B \text{ is the midpoint of } PR[/tex]                        Given

[tex]4)\text{ } PB=BR[/tex]                                               Definition of Midpoint

[tex]5)\text{ } \angle CRB\cong \angle QPB[/tex]                                    Alternate Interior Angles Theorem

[tex]6)\text{ } \Delta CRB\cong \Delta QPB[/tex]                                    AAS-Congruence

[tex]7)\text{ } QB\cong CB[/tex]                                               CPCTC

[tex]8)\text{ }A\text{ is the midpoint of } SQ[/tex]                         Given

[tex]9)\text{ } SA=AQ[/tex]                                                Definition of Midpoint

[tex]10)\text{ } AB\parallel SR[/tex]                                               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:

[tex]1)\text{ } AB\parallel SR[/tex]                                               Given

[tex]2)\text{ } \angle QSR\cong\angle QAB[/tex]                                   Corresponding Angles Theorem

[tex]3)\text{ } \angle SCQ\cong\angle ABQ[/tex]                                   Corresponding Angles Theorem

[tex]4)\text{ } \Delta SCQ\sim \Delta ABQ[/tex]                                  Angle-Angle Similarity

[tex]5)\text{ } \displaystyle \frac{SQ}{AQ}=\frac{SC}{AB}[/tex]                                            CSSTP

[tex]6)\text{ } A\text{ is the midpoint of } SQ[/tex]                       Given

[tex]8)\text{ } SA=AQ[/tex]                                              Definition of Midpoint

[tex]9)\text{ } SQ=SA+AQ[/tex]                                    Segment Addition

[tex]10)\text{ } SQ=2AQ[/tex]                                          Substitution

[tex]11)\text{ } \displaystyle \frac{2AQ}{AQ}=\frac{SC}{AB}[/tex]                                        Substitution

[tex]12)\text{ } \displaystyle 2=\frac{SC}{AB}[/tex]                                               Division Property of Equality

[tex]13)\text{ } 2AB=SC[/tex]                                          Multiplication Property of Equality

[tex]14)\text{ } SR=SC+CR[/tex]                                   Segment Addition

[tex]15)\text{ } \Delta CRB\cong\Delta QPB[/tex]                                Given

[tex]16)\text{ } CR\cong PQ[/tex]                                            CPCTC

[tex]17)\text{ }SR=SC+PQ[/tex]                                   Substitution

[tex]18)\text{ } SC=SR-PQ[/tex]                                   Subtraction Property of Equality

[tex]19)\text{ } 2AB=SR-PQ[/tex]                                 Substitution

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

Answer:

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

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