Question

can you solve it please we are to solve it till tomorrow

Answer 1

Hey there !

given :- ABCD is a parallelogram and AE = CF .

To Show :- OE = OF

construction :- join Bd and meet AC at point o .

proof :- since , diagonal of a parallelogram bisect each other ,

therefore , OA = OC
and , OD = OB

now , oa = oc
and , ae = cf

=> oa-ae = oc-cf

=> oe = of

thus BFDE is a quadrilateral whose diagonal bisect each other .
Hence , BFDE is a parallelogram . hence proved .

2). ABCD is a square

BC = CD

Given QC = RD

∴ BC - QC = CD - RD

⇒ BQ = CR

now , In ∆PBQ and ∆QCR

PB = QC  [given]

BQ = CR  [Proved above ]

∠PBQ = ∠QCR  [Each is 90°]

∴ ∆PBQ and ∆QCR [SAS Congruency]

⇒ PQ = QR, ∠BPQ = ∠CQR, ∠BQP = ∠CRQ  [C.P.C.T.]

now ,

BQC is a straight line

∴ ∠BQP + ∠PQR + ∠CQR = 180°

⇒∠BQP + ∠PQR + ∠BPQ = 180°

⇒(∠BQP + ∠BPQ) + ∠PQR = 180°

⇒180° - ∠PBQ + ∠PQR = 180° [Angle sum property for ∆PBQ]

⇒180° - 90° + ∠PQR = 180°

⇒∠PQR = 90°

now ,

∆PQR is an isosceles right angled triangle

∴ ∠QPR = 45° hence , proved _/\_

hope this helps ,

sorry i can't solve 3) right now .

be brainly together we go far ♥