show that in a quadrilateral abcd , ab+bc+cd+da>ac+bd .
In any triangle , the sum of any two sides > the third side, so
in triangle abd,
ab +ad > bd **
in triangle bcd,
bc+dc > bd ***
in triangle adc
ad + cd > ac ****
in triangle abc
ab+bc > bd ****
add up *, **, ***, and ****
2ab + 2bc + 2dc + 2ad > 2bd + 2ac
divid by 2:
ab + bc + dc + ad > bd + ac , as needed
add ** and ***
ab+bc+ad+cd >
In the quadrilateral ABCD, AB is represented by the equation y=1 where -1 < x < 1, DC is represented by the equation y = -1 where -1 < x < 2.
1) What type of quadrilateral is this? Use the equations to find coordinates for A B C and D.
2) What is the perimeter?
3) What is the area?
To prove that in a quadrilateral ABCD, the sum of its four sides (AB, BC, CD, and DA) is greater than the sum of its diagonals (AC and BD), we can use the triangle inequality theorem.
The triangle inequality theorem states that for any triangle with sides a, b, and c, the sum of any two sides of the triangle is always greater than the third side.
Now, let's apply this theorem to the quadrilateral ABCD.
In triangle ABC, we have the sides AB, BC, and AC.
Applying the triangle inequality theorem to this triangle, we get:
AB + BC > AC (1)
In triangle CDA, we have the sides CD, DA, and AC.
Applying the triangle inequality theorem to this triangle, we get:
CD + DA > AC (2)
Now, if we add inequalities (1) and (2), we get:
AB + BC + CD + DA > AC + AC
AB + BC + CD + DA > 2AC
Similarly, let's apply the triangle inequality theorem to triangles ABD and BCD.
In triangle ABD, we have the sides AB, BD, and AD.
Applying the triangle inequality theorem to this triangle, we get:
AB + BD > AD (3)
In triangle BCD, we have the sides BC, CD, and BD.
Applying the triangle inequality theorem to this triangle, we get:
BC + CD > BD (4)
Adding inequalities (3) and (4), we get:
AB + BD + BC + CD > AD + BD
AB + BC + CD + DA > 2BD
Since AC is one of the diagonals of the quadrilateral ABCD and BD is another diagonal, we can write the inequality as:
AB + BC + CD + DA > 2AC + 2BD
Finally, we can divide both sides of the inequality by 2 to get rid of the factor of 2:
AB + BC + CD + DA > AC + BD
Therefore, we have proved that in a quadrilateral ABCD, the sum of its four sides is greater than the sum of its diagonals: AB + BC + CD + DA > AC + BD.