Question

In the above figure, PQS=SRQ,QPR=SQR .U is the point on qr such that QPU=30 then QUP is​

Answer 1

Given : In ||gm ABCD,

E a mid point of BC

DE is joined and produced to meet AB produced at F

To prove : AF = 2AB

Proof : In ∆CDE and ∆EBF

∠DEC = ∠BEF (vertically opposite angles)

CE = EB (E is mid point of BC)

∠DCE = ∠EBF (alternate angles)

∴ ∆CDE ≅ ∆EBF (SAS Axiom)

∴ DC = BF (c.p.c.t.)

But AB = DC (opposite sides of a ||gm)

∴ AB = BF

Now, AF = AB + BF = AB + AB = 2AB

Hence AF = 2AB