In the given figure, PQRS is a square and
AB||QS. M is the mid-point of AB. Prove
that PM bisects 2QPS.
Answers
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
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