In the figure, prove that:
$CD + DA + AB > BC$

"\n

To do:

We have to prove that $CD + DA + AB > BC$.

Solution:

In the given figure,

$ABCD$ is a quadrilateral and $AC$ is joined.

In $\triangle ACD$,

$CD + DA > CA$          (Sum of two sides of a triangle is greater than its third side)

Adding $AB$ to both sides, we get,

$CD + DA + AB > CA + AB$

In $\triangle ABC$,

$CA + AB > BC$

This implies,

$CD + DA + AB > BC$. 

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Updated on: 10-Oct-2022

BC$.Solution:In the given figure,$ABCD$ is a quadrilateral and $AC$ is joined.In $triangle ACD$,$CD + DA > CA$          (Sum of two sides of a triangle is greater than its third side)Adding $AB$ to both sides, we get,$CD + DA + AB > CA + AB$In $tria','sharer','toolbar=0,status=0,width=580,height=325');" href="javascript: void(0)" title="Share on Facebook"> BC$.Solution:In the given figure,$ABCD$ is a quadrilateral and $AC$ is joined.In $triangle ACD$,$CD + DA > CA$          (Sum of two sides of a triangle is greater than its third side)Adding $AB$ to both sides, we get,$CD + DA + AB > CA + AB$In $tria','sharer','toolbar=0,status=0,width=580,height=325');" href="javascript: void(0)" title="Share on LinkedIn">

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