The diagonals that a quadrilateral deserve to determine whether it is a parallelogram, a rectangle, a rhombus, etc.. We will certainly list and prove the key theorems here.

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Theorem 1: If the diagonals that a quadrilateral bisect each other then the square is a parallelogram.

**Proof:** allow the square be ABCD through diagonals AC and BD intersecting in ~ P:

Then AP = PC and DP = PB because the diagonals bisect each other. Now consider triangles APD and also CPB. The vertical angle APD and also CPB room equal, and we have actually pairs of sides that room equal (AP = PC and also DP = PB), so this triangles space congruent through **SAS**. Together a consequence, the matching angles, DAP and BCP are equal. Yet these are alternating interior angle for lines ad and BC v transversal AC. Therefore this proves side advertisement is parallel to side BC.

Similarly, if we take into consideration triangles APB and also CPD, they room congruent through the exact same reasoning, so equivalent angles BAP and DCP space equal. These are alternating interior angle for lines ab and DC, therefore those present are likewise parallel.

Therefore ABCD is a parallelogram, through definition.

**Theorem 2:** If the diagonals the a quadrilateral bisect every other and also have the exact same length, climate the quadrilateral is a rectangle.

**Proof:** due to the fact that the diagonals bisect every other, we currently know (from organize 1) the it is a parallelogram, so every we have to prove is that the angles at the vertices are ideal angles.

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

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

**Proof:** Again allow the square be ABCD v diagonals AC and BD intersecting in ~ P:

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

They space all congruent through **SAS**. Because that example, triangle APB is congruent to triangle CPB since they re-superstructure a common side BD, sides AP and CP space congruent (since ns is the midpoint that AC), and also the had angles space both appropriate angles. Since they space all congruent, their 3rd sides (the hypotenuse of each) space congruent (CPCTC). A rhombus is a quadrilateral in which all 4 sides space congruent, so ABCD is a rhombus.

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

That is, if the quadrilateral is ABCD through diagonals AC and also BC intersecting at P, and if AP = CP, then abdominal muscle = BC and advertisement = BD:

**Proof:** In this case, triangles APB and also CPB room congruent (by **SAS**), and triangles APD and also CPD room congruent. Therefore ab = BC and ad = CD.

Narrowing the Type

When a quadrilateral is recognized to it is in of certain special type, then added properties of the diagonals deserve to narrow the type. The adhering to theorems show this:

**Theorem 5:** If the diagonals that a parallelogram space congruent, then the parallelogram is a rectangle.

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

Consider the overlapping triangles ADC and also BCD. Because opposite sides of a parallelogram are congruent, advertisement = BC. Due to the fact that the diagonals of the parallelogram room congruent, AC = BD, and also the overlapping triangles have actually a typical side, DC. Because of this they room congruent by **SSS**. So angle ADC and BCD are congruent. But these room same-side inner angles for parallel lines ad and BC through transversal DC. Since same-side inner angles add to 180o, each need to be 90o, for this reason the parallel is a rectangle.

This to organize is regularly used through carpenters to examine a door or home window to watch if the is really rectangular. Very first the carpenter steps the opposite sides. If they are the very same length, then he steps the diagonals. If lock too room the same length, then he knows the angle are ideal angles.

**Theorem 6:** If the diagonals that a parallelogram space perpendicular, then the parallelogram is a rhombus.

**Proof:** permit the parallel be ABCD with perpendicular diagonals AC and also BD intersecting at P:

The diagonals the a parallel bisect every other, so triangle APB and also CPB room congruent through **SAS**. Because of this the equivalent parts, sides ab and CB room congruent. Likewise, triangles BPC and also DPC are congruent, so sides BC and DC room congruent, and likewise sides ad and CD space congruent. So every 4 sides are congruent, which makes the parallel a rhombus.

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There are other theorems which can be stated, however the main principles revolve roughly congruent triangles created by the diagonals and sides of the quadrilateral.