If PQRS is a rectangle and M is the midpoint of RS, prove PM is congruent to QM.

P----------------Q
|\ /|
| \ / |
| \ / |
| \ / |
| \ / |
R--------M-------S
*be as explanatory as possible

Well, if it is a rhombus we know all the sides are of equal length AND opposite sides are parallel.

All we have to prove is that the corner angles are equal
CBE=BCE given
so
CBE=BCE=ADE=DAE intersect parallel lines
BE = CE isosceles
AD = BC rhombus
so triangle AED congruent triangle BEC from angle side angle
triangle ABE congruent to BCE side side side
so
angle ABC = angle BCD= angle CDA
4 x = 360
x = 90 degrees

P(-3;7),Q(5;13),R(8;9) and S(0;3) are the vertices of a quadrilateral in a cartesian plane.

Prove this P(-3;7),Q(5;13),R(8;9) and S(0;3) are the vertices of a quadrilateral in a Cartesian plane.

To prove that PM is congruent to QM, we need to use the properties of rectangles and midpoints.

1. First, we know that PQRS is a rectangle. In a rectangle, opposite sides are congruent. Therefore, QR is congruent to PS and PQ is congruent to RS.

2. Next, we are given that M is the midpoint of RS. This means that M divides RS into two congruent segments, RM and MS.

3. Now, let's consider the triangle PMQ. We want to prove that PM is congruent to QM.

4. By using the properties of rectangles and midpoints, we can argue that triangle PMQ is congruent to triangle SMQ. Here's why:
- We have already established that PQ is congruent to RS (property of rectangles).
- We know that M is the midpoint of RS, so RM is congruent to MS (property of a midpoint).
- And since QR is congruent to PS, we can say that PMQ is congruent to SMQ by the side-side-side (SSS) congruence criterion.

5. Now, since triangle PMQ is congruent to triangle SMQ, corresponding parts of congruent triangles are congruent. Therefore, PM is congruent to QM by the corresponding parts of congruent triangles (CPCTC) theorem.

In conclusion, we have proven that PM is congruent to QM using the properties of rectangles and midpoints, as well as the congruence criterion and theorem.