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