Question:
In a quadrilateral $\mathrm{ABCD}_{1}$ show that
$(A B+B C+C D+D A)<2(B D+A C)$
Solution:
Given: Quadrilateral ABCD
To prove: (AB + BC + CD + DA) < 2(BD + AC).
Proof:
In ∆AOB,
$O A+O B>A B \quad \ldots \ldots(i)$
In ∆BOC,
$O B+O C>B C \quad \ldots \ldots(i i)$
In ∆COD,
$O C+O D>C D \quad \ldots \ldots($ iii $)$
In ∆AOD,
$O D+O A>A D \quad \ldots \ldots(i v)$
Adding $(i),(i i),(i i i)$ and $(i v)$, we get
$2(O A+O B+O C+O D)>(A B+B C+C D+D A)$
$\Rightarrow 2(O B+O D+O A+O C)>(A B+B C+C D+D A)$
$\Rightarrow 2(B D+A C)>(A B+B C+C D+D A)$
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