Show that ab + bc +cd + da is greater than 2ac
Answers
Answered by
0
When you draw a diagonal across the quadrilateral abcd, two triangles will be formed.
AC is the diagonal.
We know that the sum of 2 sides is greater than the third side.
Therefore,
ab+bc>ac
AND
cd+da>ac.
We can add these two, so we will get
ab+bc+cd+da>ac+ac
Therefore ab+bc+cd+da>2ac.
Hence proven.
AC is the diagonal.
We know that the sum of 2 sides is greater than the third side.
Therefore,
ab+bc>ac
AND
cd+da>ac.
We can add these two, so we will get
ab+bc+cd+da>ac+ac
Therefore ab+bc+cd+da>2ac.
Hence proven.
Similar questions