**Diagonals in Quadrilaterals**

Main Theorems

The diagonals of a quadrilateral can determine whether it is a parallelogram, a rectangle, a rhombus, etc.. We will list and prove the main theorems here.

**Theorem 1:** If the diagonals of a quadrilateral bisect each other then the quadrilateral is a parallelogram.

**Proof: ** Let the quadrilateral be *ABCD* with diagonals *AC* and *BD* intersecting at *P*:

Then *AP* = *PC* and *DP* = *PB* since the diagonals bisect each other. Now consider triangles *APD* and *CPB*. The vertical angles *APD* and *CPB* are equal, and we have pairs of sides that are equal (*AP* = *PC* and *DP* = *PB*), so these triangles are congruent by **SAS**. As a consequence, the corresponding angles, *DAP* and *BCP* are equal. But these are alternate interior angles for lines *AD* and *BC* with transversal *AC*. So this proves side *AD* is parallel to side *BC*.

Similarly, if we consider triangles *APB* and *CPD*, they are congruent by the same reasoning, so corresponding angles *BAP* and *DCP* are equal. These are alternate interior angles for lines *AB* and *DC*, so those lines are also parallel.

Therefore *ABCD* is a parallelogram, by definition.

**Theorem 2:** If the diagonals of a quadrilateral bisect each other and have the same length, then the quadrilateral is a rectangle.

**Proof:** Since the diagonals bisect each other, we already know (from Theorem 1) that it is a parallelogram, so all we need to prove is that the angles at the vertices are right angles.

Again let the quadrilateral be *ABCD* with diagonals *AC* and *BD* intersecting at *P*. Since the diagonals bisect each other, *P* is the midpoint of both diagonals. That is, *AP* = *PB* and *CP* = *PD*. But the diagonals are also of the same length, so *AP* + *PB* = *CP* + *PD*, and by substitution this gives us *AP* + *AP* = *CP* + *CP*, or 2*AP*= 2*CP*. That is, *AP* = *CP*. Likewise, *PB* = *PD*. Consequently, all 4 triangles *APD*, *APB*, *CPD*, and *BPC* are isosceles. So angles *PAD* and *PDA* are congruent, angles *PBC* and *PCB* are congruent, angles *PAB* and *PBA* are congruent, and angles *PDC* and *PCD* are congruent. We already saw in the proof of Theorem 1 that triangles *APD* and *CPB* are congruent, as are triangles *APB* and *CPD*. By angle addition, it follows that the 4 angles of the quadrilateral (angles *ABC*, *BCD*, *CDA*, and *DAB*) are all equal. But the angles of a quadrilateral add to 360^{o}, and therefore each of these 4 angles must be 90^{o}.

**Theorem 3:** If the diagonals of a quadrilateral bisect each other and are perpendicular then the quadrilateral is a rhombus.

**Proof:** Again let the quadrilateral be *ABCD* with diagonals *AC* and *BD* intersecting at *P*:

Since they bisect each other and are perpendicular, triangles *APB*, *BPC*, *CPD*, and *DPA* are right triangles.

They are all congruent by **SAS**. For example, triangle *APB* is congruent to triangle *CPB* because they share a common side *BD*, sides *AP* and *CP* are congruent (since *P* is the midpoint of *AC*), and the included angles are both right angles. Since they are all congruent, their third sides (the hypotenuse of each) are congruent (CPCTC). A rhombus is a quadrilateral in which all four sides are congruent, so *ABCD* is a rhombus.

**Theorem 4: ** If the diagonals of a quadrilateral are perpendicular and one bisects the other, then the quadrilateral is a kite.

That is, if the quadrilateral is *ABCD* with diagonals *AC* and *BC* intersecting at *P*, and if *AP* = *CP*, then *AB* = *BC* and *AD* = *BD*:

**Proof:** In this case, triangles *APB* and *CPB* are congruent (by **SAS**), and triangles *APD* and *CPD* are congruent. Therefore *AB* = *BC* and *AD* = *CD*.

Narrowing the Type

When a quadrilateral is known to be of certain special type, then additional properties of the diagonals can narrow the type. The following theorems demonstrate this:

**Theorem 5:** If the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle.

**Proof:** Let the parallelogram be *ABCD* with congruent diagonals *AC* and *BD*.

Consider the overlapping triangles *ADC* and *BCD*. Since opposite sides of a parallelogram are congruent, *AD* = *BC*. Since the diagonals of the parallelogram are congruent, *AC* = *BD*, and the overlapping triangles have a common side, *DC*. Therefore they are congruent by **SSS**. So angle *ADC* and *BCD* are congruent. But these are same-side interior angles for parallel lines *AD* and *BC* with transversal *DC*. Since same-side interior angles add to 180^{o}, each must be 90^{o}, so the parallelogram is a rectangle.

This theorem is often used by carpenters to check a door or window to see if it is really rectangular. First the carpenter measures the opposite sides. If they are the same length, then he measures the diagonals. If they too are the same length, then he knows the angles are right angles.

**Theorem 6:** If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.

**Proof:** Let the parallelogram be *ABCD* with perpendicular diagonals *AC* and *BD* intersecting at *P*:

The diagonals of a parallelogram bisect each other, so triangles *APB* and *CPB* are congruent by **SAS**. Therefore the corresponding parts, sides *AB* and *CB* are congruent. Likewise, triangles *BPC* and *DPC* are congruent, so sides *BC* and *DC* are congruent, and similarly sides *AD* and *CD* are congruent. So all 4 sides are congruent, which makes the parallelogram a rhombus.

There are other theorems which could be stated, but the main ideas revolve around congruent triangles formed by the diagonals and sides of the quadrilateral.