Question

B'
26.) ABCD is a parallelogram and E is the
point on CD, such that CE = 2ED. AE is
joined meeting BD in F and BC produced
in G. Prove that AG=4AF​

Answer 1

Solution :-

→ CE = 2ED

→ CE + ED = CD

→ 2ED + ED = CD

→ 3ED = CD = BA { Opposite sides of ll gm are equal. } -------- Eqn.(1)

now, in ∆DEF and ∆BAF we have,

→ ∠FDE = ∠FBA { since DC || AB , alternate angles. }

→ ∠FED = ∠FAB

so,

→ ∆DEF ~ ∆BAF

then,

→ EF / AF = DE/BA

→ EF / AF = DE/3ED

→ EF / AF = 1/3

→ EF = (AF/3) ------- Eqn.(2)

now, in ∆GAB , we have,

→ EC || AB { Opposite sides of ll are parallel}

so,

→ GE / GA = EC / AB { By BPT. }

→ GE / GA = (DC - DE) / AB

→ GE / GA = 3DE - DE / 3DE { using Eqn.(1) }

→ GE / GA = 2DE/3DE

→ GE / GA = 2/3

→ GE = (2GA)/3 ------------ Eqn.(3)

now,

→ GE = GA - AE

→ GE = GA - (AF + EF)

putting value of Eqn.(2) and Eqn.(3)

→ (2GA/3) = GA - (AF + AF/3)

→ AF + AF/3 = GA - 2GA/3

→ (3AF + AF)/3 = (3GA - 2GA)/3

→ 4AF = GA

→ AG = 4AF (Proved).

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