Question
ABCD is a quadrilateral.
$
\text { Is } A B+B C+C D+D A>A C+B D \text { ? }
$
Image

Answer

Yes, given $A B C D$ is a quadrilateral. Here, diagonal $A C$ divides the quadrilateral $A B C D$ into two $\triangle A B C$ and $\triangle A C D$.
In $\triangle A B C, \quad A B+B C>A C\quad\quad ...(i)$
In $\triangle A C D, \quad C D+D A>A C\quad\quad ...(ii)$
On adding Eqs. (i) and (ii), we get
$
A B+B C+C D+D A>2 A C \quad\quad ...(iii)$
Similarly, the diagonal $B D$ divides the quadrilateral $A B C D$ into two $\triangle B C D$ and $\triangle A B D$.
$
\begin{array}{ll}
\text { In } \triangle B C D, & B C+C D>B D \quad ...(iv)\\
\text { In } \triangle A B D, & A B+D A>B D \quad ...(v)
\end{array}
$
On adding Eqs. (iv) and (v), we get
$
(B C+C D+A B+D A)>2 B D \quad ...(vi)
$
Again, adding Eqs. (iii) and (vi), we get
$
\begin{array}{l}
A B+B C+C D+D A+B C+C D+A B+D A>2 A C+2 B D \\
\Rightarrow 2(A B+B C+C D+D A)>2(A C+B D) \\
\Rightarrow \quad A B+B C+D C+D A>A C+B D
\end{array}
$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free