Question

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Answer 1

Step-by-step explanation:

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Answer 2

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