If pqrs is a parallelogram and ab parallel PS, prove that oc parallel sr
Answers
In △ OAB, we have PS ∥ AB.Since PS ∥ AB and PA is a transversal, then∠OPS = ∠OAB [corresponding angles] ............(1) Since PS ∥ AB and SB is a transversal, then∠OSP = ∠OBA [corresponding angles]...(2) In △ OPS and △ OAB, ∠OPS = ∠OAB [using (1)] ∠OSP = ∠OBA [using (2)] ⇒ △ OPS ~ △ OAB [AA similarity] ⇒ OPOA = PSAB = OSOB [corresponding sides of similar △'s are proportional] ..............(3) Since, PQRS is a ∥gm, so QR ∥ AB.Since QR ∥ AB and QA is a transversal,∠CQR = ∠CAB [corresponding angles] ............(4) Since QR ∥ AB and RB is a transversal,∠CRQ = ∠CBA [corresponding angles] ............(5)
In △ CQR and △ CAB ∠CQR = ∠CAB [using (4)] ∠CRQ = ∠CBA [using (5)] ⇒ △ CQR ~ △ CAB [AA similarity] ⇒ CQCA = QRAB = CRCB ⇒ CQCA = PSAB = CRCB [as, QR = PS].......(6) From (3) & (6) OPOA = CQCA ⇒ AOOP = ACQC ⇒ AOOP − 1 = ACQC − 1 ⇒ AO − OPOP = AC − QCQC ⇒ APOP = AQQC ..............(7)
Now, In △ AOC, we have, APOP = AQQC [using (7)]⇒ PQ ∥ OC [converse of Thales theorem] Now, PQ ∥ SR as PQRS is a ∥gm. So, OC ∥ SR.
This can be proved by applying similarity of triangles and converse of Thales Theorem .
Step-by-step explanation:
To prove - OC║SR
Proof - In ΔOPS and ΔOAB
∠POS = ∠AOB (common in both)
∠OSP = ∠OBA (corresponding angles are equal as PS║AB)
=> ΔOPS ~ ΔOAB [AA criteria]
=> PS/AB = OS/OB ........................(1) (sides in similar triangles are proportional)
In ΔCAB and ΔCRQ
As, QR║AB
=> ∠QCR = ∠ACB (common)
=> ∠CBA = ∠CRQ (corresponding angles are equal)
=> ΔCAB ~ ΔCQR [AA criteria]
=> CR/CB = QR/AB (sides in similar triangles are proportional)
Also, PS = QR [ PQRS is parallelogram]
=> CR/CB = PS/AB ......................(2)
From (1) and (2)
=> OS/OB = CR/CB
=> OB/OS = CB/CR
Subtracting 1 from both sides
So, OB/OS - 1 = CB/CR - 1
=> (OB - OS)/OS = (CB - CR)/CR
=> BS/OS = BR/CR
By converse of Thales Theorem
=> OC║SR . Hence proved .
........Brainliest..................