Prove that Quadrilateral ABCD is a Rhombus
show all the work
![Prove that Quadrilateral ABCD is a Rhombus show all the work class=](https://us-static.z-dn.net/files/da9/510ad2cab2bcf9ca70e1aeb3febb3589.png)
Answer:
Say that the shape is a parallelogram with equal length sides;
AB = CD
ABD = CDB as the bisecting lines (you have drawn) prove ABD : CDB are both isosceles triangles.
Therefore ABD = ADB
Draw the shape's diagonals as each others' perpendicular bisectors;
Draw the shape's diagonals bisect both pairs of opposite angles.
If two non congruent angles were given in example 70 degree and 110 degree then say M ∴ = (1/2 )(M∴BAD) = 35 = (1/2 )(M∴BEF) = 35
and do the same for 1/2(M∴ABC) = 55 = 1/2 (M∴CDA) = 55