I"https://snucongo.org/if-abcd-is-a-parallelogram-which-statement-would-prove-that-abcd-is-a-rhombus/imager_2_7764_700.jpgm reasoning of a parallelogram whose diagonals are perpendicular. Name the parallelogram.
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If girlfriend guessed that it was a square, then you didn"https://snucongo.org/if-abcd-is-a-parallelogram-which-statement-would-prove-that-abcd-is-a-rhombus/imager_2_7764_700.jpgt read the heading for this section an extremely well. It"https://snucongo.org/if-abcd-is-a-parallelogram-which-statement-would-prove-that-abcd-is-a-rhombus/imager_2_7764_700.jpgs a rhombus! The pretty thing around working v parallelograms is that the diagonals create lots that triangles just begging to be proven congruent. In figure 16.6, parallelogram ABCD has actually perpendicular diagonals. The congruent triangles are trying to connect with you. Listen closely.
Figure 16.6Parallelogram ABCD with AC ? BD.
Theorem 16.6: If the diagonals the a parallelogram room perpendicular, the parallel is a rhombus.
Let"https://snucongo.org/if-abcd-is-a-parallelogram-which-statement-would-prove-that-abcd-is-a-rhombus/imager_2_7764_700.jpgs simply jump right right into the video game plan. You understand that AC ? BD, for this reason m?AMB = 90 and m?CMB = 90. Since the diagonals of a parallelogram bisect each other, you understand that to be ~= MC. The reflexive building of ~= permits you to write BM ~= BM. By the SAS Postulate, you recognize that ?AMB ~= ?CMB. By CPOCTAC, you recognize that ab ~= BC. Because abdominal ~= BC and ab ~= BC are nearby sides, you have a parallelogram v congruent nearby sides, a.k.a. A rhombus.
|1.||Parallelogram ABCD has actually AC ? BD||Given|
|2.||?AMB and also ?CMB are right||Definition of ?|
|3.||m?AMB = 90 and also m?CMB = 90||meaning of appropriate angle|
|4.||?AMB ~= ?CMB||meaning of ~=|
|5.||AM ~= MC||Theorem 15.6|
|6.||BM ~= BM||Reflexive residential property of ~=|
|7.||?AMB ~= ?CMB||SAS Postulate|
|8.||AB ~= BC||CPOCTAC|
|9.||Parallelogram ABCD is a rhombus||Definition that rhombus|
Now let"https://snucongo.org/if-abcd-is-a-parallelogram-which-statement-would-prove-that-abcd-is-a-rhombus/imager_2_7764_700.jpgs gain a little tricky. Intend you have a rectangle ABCD. Find the midpoints of each of the political parties of the rectangle and also join them together consecutively to kind the quadrilateral MNOP, as presented in number 16.7. What sort of square is made?
Figure 16.7Rectangle ABCD, through midpoints of every side joined together consecutively to type the quadrilateral MNOP.
From the picture, the looks kind of like a parallelogram. You need to be careful, though, due to the fact that looks can be deceiving. It additionally looks choose the diagonals that the newly created quadrilateral room perpendicular. If the drawing is accurate, you could be tempted to conclude that the square is a rhombus. Let"https://snucongo.org/if-abcd-is-a-parallelogram-which-statement-would-prove-that-abcd-is-a-rhombus/imager_2_7764_700.jpgs prove it.Theorem 16.7: If the midpoints the the sides of a rectangle are joined in order, the quadrilateral formed is a rhombus.
You require a severe game setup for this one. Because M, N, O, and P space the midpoints of AB, BC, CD, and also AD, you understand that BN ~= NC ~= AP ~= PD and AM ~= MB ~= OD ~= CO. Due to the fact that you"https://snucongo.org/if-abcd-is-a-parallelogram-which-statement-would-prove-that-abcd-is-a-rhombus/imager_2_7764_700.jpgre managing a rectangle, you know that m?A = 90, m?B = 90 , m?C = 90, and also m?D = 90. Therefore by the SAS Postulate, ?PAM ~= ?NBM ~= ?PDO ~= ?NCO. Using the CPOCTAC principal MN ~= MP ~= PO ~= NO. For this reason opposite sides space congruent and also quadrilateral MNOP is a parallelogram. Also, nearby sides are congruent, so parallelogram MNOP is a rhombus.
|1.||Rectangle ABCD, with midpoints of every side joined together consecutively to type the quadrilateral MNOP||Given|
|2.||BC ~= NC, AP ~= PD, and also OD ~= CO||Definition the midpoint|
|3.||ab ~= CD and also BC ~= AD||Theorem 15.4|
|4.||BN = 1/2BC, AP = 1/2AD, am = 1/2AB and CO = 1/2CD||Theorem 9.1|
|5.||BN = NC = AP = PD and also AM = MB = OD = CO||Substitution (steps 2, 3, 4)|
|6.||BN ~= NC ~= AP ~= PD and AM ~= MB ~= OD ~= CO||Definition of ~=|
|7.||m?A = 90. M?B = 90 , m?C = 90 and also m?D = 90||Definition that rectangle|
|8.||?PAM ~= ?NBM ~= ?PDO ~= ?NCO||SAS Postulate|
|9.||MN ~= MP ~= PO ~= NO||CPOCTAC|
|10.||Quadrilateral MNOP is a parallelogram||Theorem 16.2|
|11.||Quadrilateral MNOP is a rhombus||Definition the rhombus|
Excerpted native The complete Idiot"https://snucongo.org/if-abcd-is-a-parallelogram-which-statement-would-prove-that-abcd-is-a-rhombus/imager_2_7764_700.jpgs overview to Geometry 2004 by Denise Szecsei, Ph.D.. All legal rights reserved including the right of reproduction in totality or in component in any form. Provided by plan with Alpha Books, a member that Penguin group (USA) Inc.
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