How do you prove that the diagonals in a rhombus are perpendicular to each other?

Answer 1

I will outline a way to prove it for you. I will also five a simple vector proof for those that have studied vectors. For the first proof, one can often cite some of these as known facts or refer to theorems in a text. 1. First show that a rhombus is a parallelogram 2. Next, using the above, show that diagonals of the rhombus divide it into 4 congruent triangles. 3. Last, use CPCTC and not that all 4 middle angles are congruent so that are 90 degrees. From this is is easy to say that the diagonals are perpendicular. Hints. to prove 1, use the fact that all 4 sides of the rhombus are congruent and then use SSS to find two congruent triangles. Then use CPCTC to show that the angles are the same and find a transversal. Look at same side interior angles cut by that transversal and say something about them being parallel. 2. Use SSS again and find 4 congruent triangles and look at the diagonals. I will help more by giving you another proof using vectors that is really much more straightforward. A rhombus is a quadrilateral with all sides having equal length. This means that if two vectors, a and b that form the corner of a rhombus, then the magnitude of a and b are equal The diagonals of the parallelogram are precisely a+b and a-b. Now look at the dot product of a+b and a-b and see that it is zero and remember that a dot product of zero means the vectors are perpendicular or orthogonal The first part is a pure synthetic geometry approach and if anyone need more help to finish that, just ask, The second part is a vector proof which is elegant because it is so simple.