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 2AP= 2CP.  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 360o, and therefore each of these 4 angles must be 90o.

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 180o, each must be 90o, 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.