Question

In the given figure, PQRS is a square and

AB||QS. M is the mid-point of AB. Prove

that PM bisects 2QPS.​

Answer 1

Answer:

Given PQRS is a square and M is the midpoint of PQ

Also, RM⊥AB

In △APM & △BMQ, we have:

PM=MQ (M is the midpoint of PQ)

∠APM=∠BQM=90°

∠AMP=∠BMQ (Vertically opposite angles)

By ASA congruence axion,

△APM≅△BMQ

∴AM=MB →1

Consider right angled △RMA

RA

2

=AM

2

+RM

2

(By Pythagoras theorem)

⇒RA

2

=MB

2

+RM

2

→2 (From 1)

Similarly in right angled △RMB

RB

2

=MB

2

+RM

2

(By Pythagoras theorem) →3

From 2 and 3 we get,

RA

2

=BR

2

⇒AR=BR

Answer 2

Step-by-step explanation:

Given -

• PQRS is a square
• AB || QS
• M is the mid-point of AB

To Prove - PM bisects ∠QPS

Proof -

∠SQA = ∠BAR (Corresponding Angles)

∠QSB = ∠ABR (Corresponding Angles)

∠SQA = ∠QSB (Diagonals of a square bisect each other)

Therefore, ∠BAR = ∠ABR

Therefore, AR = BR (Sides opposite to equal angles are equal)

QR=SR (Sides of a square)

QA+AR=SB+BR

QA=SB (Since, AR = BR)

In △PQA and △PSB

PQ = PS (sides of a square)

∠PQA = ∠PSB (Angles of a square)

QA=SB (Proved above)

Therefore by SAS congruence criteria, △PQA ≅ △PSB

By CPCT,

PA = PB

∠QPA = ∠SPB ----- (1)

In △PAM and △PBM

PA = PB (by CPCT)

AM = BM (M is the mid-point of AB)

PM = PM (Common)

Therefore, by SSS congruence criteria, △PAM ≅ △PBM

By CPCT,

∠APM = ∠BPM ------ (2)

Adding (1) and (2)

∠QPA + ∠APM = SPB + BPM

∠QPM = ∠SPM

Therefore, we can say that PM bisects ∠QPS

HENCE PROVED