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Step-by-step explanation:
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Answer:
Step-by-step explanation:
→ (1) we have
Δ ABC ~ ΔFEG .....(1)
⇒ 2A=2F
AND ∠C=∠G
⇒ 1/2 ∠C= 1/2 ∠G
⇒ ∠1=∠3 AND ∠2=∠4 ......(2)
[∴ CD AND GH ARE BISECTORS OF ∠C AND ∠G RESPECTIVELY.]
THUS, IN Δ"S ACD AND FGH , WE HAVE
∠A=∠F (FROM 1)
∠2=∠4 (FROM 2)
∴ BY AA- CRITERION OF SIMILARITY , WE HAVE
ΔACD ~ ΔFGH OR
ΔDCA ~ ΔHGF
(2) WE HAVE ,
ΔACD ~ ΔFGH
⇒ AC/FG = CD/GH
(3) WE HAVE IN Δ"S DCB AND HGE
∠1=∠3 (FROM 2)
⇒ ∠B=∠E
(∵ ΔABC ~ ΔFEG)
THUS , BY AA-CRITERION OF SIMILARITY WE HAVE
ΔDCB ~ ΔHGF