1) In quadrilateral ABCD, Prove that AB + BC + CD + DA
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Answer:
2AC + 2DB ∠ 2AB + 2BC + 2CD + 2DA
= AC+DB ∠ AB + BC + CD +DA
Step-by-step explanation:
By In equality theorem, sum of any two
sides of Δ is grater than third one
In Δ ABC
AC ∠ AB+BC ... (1)
In ΔBCD
DB ∠ BC +CD ... (2)
In ΔACD
AC ∠ DA +DC ... (3)
In Δ ADB
DB ∠ AD + AB ... (4)
adding (1),(2),(3),(4)
= 2AC + 2DB ∠ 2AB + 2BC + 2CD + 2DA
= AC+DB ∠ AB + BC + CD +DA
Hence proved.
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