Question

pls help me in this question...no spams pls!!​

Answer 1

Step-by-step explanation:

(i)

In ΔBDF and ΔBCE,

DF ║ CE

∠BDF = ∠BCE

∠BFD = ∠BEC

∴ ΔBDF ~ ΔBCE{AAA theorem}

⇒ (BD/BC) = (BF/BE) {BPT}

⇒ (BD/2BD) = (BF/BE)

⇒ 2BF = BE

⇒ BF = BE

(ii)

In ΔAFD and ΔAEG,

FD ║ EG.

⇒ ∠DFA = ∠GEA

⇒ ∠FDA = ∠EGA

∴ ΔAFD ~ ΔAEG

⇒ (AG/GD) = (AE/FE)

Given AE = EB

⇒ AE = 2FB

⇒ (AE/FB) = 2

⇒ (AG/GD) = (2/1)

⇒ AG : GD = 2 : 1

Hope it helps!