The Diagonals Of A Quadrilateral
Diagonals in Quadrilaterals
Main Theorems
The diagonals of a quadrilateral can determine whether it is a parallelogram, a rectangle, a rhomb, etc.. We volition listing and prove the master theorems here.
Theorem 1: If the diagonals of a quadrilateral bifurcate each other then the quadrilateral is a parallelogram.
Proof: Allow the quadrilateral be ABCD with diagonals Air-conditioning and BD intersecting at P:
Then AP = PC and DP = Lead since the diagonals bifurcate 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), and then these triangles are congruent by SAS. As a consequence, the respective 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, and then corresponding angles BAP and DCP are equal. These are alternating interior angles for lines AB and DC, so those lines are also parallel.
Therefore ABCD is a parallelogram, past definition.
Theorem 2: If the diagonals of a quadrilateral bisect each other and have the same length, so the quadrilateral is a rectangle.
Proof: Since the diagonals bifurcate each other, we already know (from Theorem 1) that it is a parallelogram, and so all we need to testify is that the angles at the vertices are correct angles.
Over again let the quadrilateral be ABCD with diagonals AC and BD intersecting at P. Since the diagonals bifurcate each other, P is the midpoint of both diagonals. That is, AP = Pb and CP = PD. But the diagonals are besides of the aforementioned length, and so AP + PB = CP + PD, and by substitution this gives us AP + AP = CP + CP, or twoAP= twoCP. That is, AP = CP. Likewise, Atomic number 82 = PD. Consequently, all 4 triangles APD, APB, CPD, and BPC are isosceles. Then 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. Past angle addition, it follows that the 4 angles of the quadrilateral (angles ABC, BCD, CDA, and DAB) are all equal. Only the angles of a quadrilateral add to 360o, and therefore each of these 4 angles must be 90o.
Theorem iii: If the diagonals of a quadrilateral bisect each other and are perpendicular so the quadrilateral is a rhombus.
Proof: Again let the quadrilateral be ABCD with diagonals AC and BD intersecting at P:
Since they bifurcate each other and are perpendicular, triangles APB, BPC, CPD, and DPA are right triangles.
They are all coinciding 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 rhomb 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, and so the quadrilateral is a kite.
That is, if the quadrilateral is ABCD with diagonals Air conditioning and BC intersecting at P, and if AP = CP, so AB = BC and Advert = BD:
Proof: In this instance, 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 sure special blazon, so additional properties of the diagonals can narrow the blazon. The following theorems demonstrate this:
Theorem 5: If the diagonals of a parallelogram are congruent, and so the parallelogram is a rectangle.
Proof: Allow the parallelogram be ABCD with coinciding diagonals Ac and BD.
Consider the overlapping triangles ADC and BCD. Since reverse sides of a parallelogram are congruent, Advertizement = BC. Since the diagonals of the parallelogram are congruent, Air conditioning = BD, and the overlapping triangles have a common side, DC. Therefore they are coinciding by SSS. Then angle ADC and BCD are congruent. But these are same-side interior angles for parallel lines Advertizement and BC with transversal DC. Since same-side interior angles add together to 180o, each must exist 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. Get-go the carpenter measures the opposite sides. If they are the aforementioned length, then he measures the diagonals. If they as well are the same length, and so he knows the angles are right angles.
Theorem six: If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.
Proof: Let the parallelogram exist 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 coinciding, so sides BC and DC are congruent, and similarly sides Advertising and CD are coinciding. So all 4 sides are coinciding, which makes the parallelogram a rhombus.
There are other theorems which could be stated, merely the main ideas revolve effectually congruent triangles formed by the diagonals and sides of the quadrilateral.
The Diagonals Of A Quadrilateral,
Source: http://ceemrr.com/Geometry1/Diagonals/Diagonals_print.html
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