G.SRT.B.5: Quadrilateral Proofs
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1 Regents Exam Questions G.SRT.B.5: Quadrilateral Proofs Name: G.SRT.B.5: Quadrilateral Proofs 1 Given that ABCD is a parallelogram, a student wrote the proof below to show that a pair of its opposite angles are congruent. 3 In the diagram below of quadrilateral ABCD, AD BC and DAE BCE. Line segments AC, DB, and FG intersect at E. Prove: AEF CEG What is the reason justifying that B D? 1) Opposite angles in a quadrilateral are congruent. ) Parallel lines have congruent corresponding angles. 3) Corresponding parts of congruent triangles are congruent. 4) Alternate interior angles in congruent triangles are congruent. 4 Given: parallelogram FLSH, diagonal FGAS, LG FS, HA FS Prove: LGS HAF Given: Quadrilateral ABCD, diagonal AFEC, AE FC, BF AC, DE AC, 1 Prove: ABCD is a parallelogram. 5 The accompanying diagram shows quadrilateral BRON, with diagonals NR and BO, which bisect each other at X. Prove: BNX ORX 1
2 Regents Exam Questions G.SRT.B.5: Quadrilateral Proofs 6 Given: Parallelogram ANDR with AW and DE bisecting NWD and REA at points W and E, respectively Name: 9 The diagram below shows square ABCD where E and F are points on BC such that BE FC, and segments AF and DE are drawn. Prove that AF DE. Prove that ANW DRE. Prove that quadrilateral AWDE is a parallelogram. 7 Given: Quadrilateral ABCD is a parallelogram with diagonals AC and BD intersecting at E 10 Given: Parallelogram DEFG, K and H are points on DE such that DGK EFH and GK and FH are drawn. Prove: AED CEB Describe a single rigid motion that maps onto CEB. AED Prove: DK EH 8 The diagram below shows rectangle ABCD with points E and F on side AB. Segments CE and DF intersect at G, and ADG BCG. Prove: AE BF 11 In quadrilateral ABCD, AB CD, AB CD, and BF and DE are perpendicular to diagonal AC at points F and E. Prove: AE CF
3 Regents Exam Questions G.SRT.B.5: Quadrilateral Proofs 1 Given: PROE is a rhombus, SEO, PEV, SPR VOR Name: 14 Isosceles trapezoid ABCD has bases DC and AB with nonparallel legs AD and BC. Segments AE, BE, CE, and DE are drawn in trapezoid ABCD such that CDE DCE, AE DE, and BE CE. Prove: SE EV 13 In the diagram of parallelogram ABCD below, BE CED, DF BFC, CE CF. Prove ADE BCE and prove AEB is an isosceles triangle. 15 Given: Quadrilateral ABCD with AB CD, AD BC, and diagonal BD is drawn Prove: BDC ABD 16 Prove that the diagonals of a parallelogram bisect each other. Prove ABCD is a rhombus. 17 A tricolored flag is made out of a rectangular piece of cloth whose corners are labeled A, B, C, and D. The colored regions are separated by two line segments, BM and CM, that meet at point M, the midpoint of side AD. Prove that the two line segments that separate the regions will always be equal in length, regardless of the size of the flag. 3
4 G.SRT.B.5: Quadrilateral Proofs Answer Section 1 ANS: 3 REF: 08108ge ANS: FE FE (Reflexive Property); AE FE FC EF (Line Segment Subtraction Theorem); AF CE (Substitution); BFA DEC (All right angles are congruent); BFA DEC (AAS); AB CD and BF DE (CPCTC); BFC DEA (All right angles are congruent); BFC DEA (SAS); AD CB (CPCTC); ABCD is a parallelogram (opposite sides of quadrilateral ABCD are congruent) REF: ge 3 ANS: Quadrilateral ABCD, AD BC and DAE BCE are given. AD BC because if two lines are cut by a transversal so that a pair of alternate interior angles are congruent, the lines are parallel. ABCD is a parallelogram because if one pair of opposite sides of a quadrilateral are both congruent and parallel, the quadrilateral is a parallelogram. AE CE because the diagonals of a parallelogram bisect each other. FEA GEC as vertical angles. AEF CEG by ASA. REF: 01138ge 4 ANS: Because FLSH is a parallelogram, FH SL. Because FLSH is a parallelogram, FH SL and since FGAS is a transversal, AFH and LSG are alternate interior angles and congruent. Therefore LGS HAF by AAS. REF: b 5 ANS: Because diagonals NR and BO bisect each other, NX RX and BX OX. BXN and OXR are congruent vertical angles. Therefore BNX ORX by SAS. REF: b 1
5 6 ANS: Parallelogram ANDR with AW and DE bisecting NWD and REA at points W and E (Given). AN RD, AR DN (Opposite sides of a parallelogram are congruent). AE = 1 AR, WD = 1 DN, so AE WD (Definition of bisect and division property of equality). AR DN (Opposite sides of a parallelogram are parallel). AWDE is a parallelogram (Definition of parallelogram). RE = 1 AR, NW = 1 DN, so RE NW (Definition of bisect and division property of equality). ED AW (Opposite sides of a parallelogram are congruent). ANW DRE (SSS). REF: geo 7 ANS: Quadrilateral ABCD is a parallelogram with diagonals AC and BD intersecting at E (Given). AD BC (Opposite sides of a parallelogram are congruent). AED CEB (Vertical angles are congruent). BC DA (Definition of parallelogram). DBC BDA (Alternate interior angles are congruent). AED CEB (AAS). 180 rotation of AED around point E. REF: geo 8 ANS: Rectangle ABCD with points E and F on side AB, segments CE and DF intersect at G, and ADG BCE are given. AD BC because opposite sides of a rectangle are congruent. A and B are right angles and congruent because all angles of a rectangle are right and congruent. ADF BCE by ASA. AF BE per CPCTC. EF FE under the Reflexive Property. AF EF BE FE using the Subtraction Property of Segments. AE BF because of the Definition of Segments. REF: ge 9 ANS: Square ABCD; E and F are points on BC such that BE FC ; AF and DE drawn (Given). AB CD (All sides of a square are congruent). ABF DCE (All angles of a square are equiangular). EF FE (Reflexive property). BE + EF FC + FE (Additive property of line segments). BF CE (Angle addition). ABF DCE (SAS). AF DE (CPCTC). REF: ge
6 10 ANS: Parallelogram DEFG, K and H are points on DE such that DGK EFH and GK and FH are drawn (given). DG EF (opposite sides of a parallelogram are congruent). DG EF (opposite sides of a parallelogram are parallel). D FEH (corresponding angles formed by parallel lines and a transversal are congruent). DGK EFH (ASA). DK EH (CPCTC). REF: ge 11 ANS: Quadrilateral ABCD, AB CD, AB CD, and BF and DE are perpendicular to diagonal AC at points F and E (given). AED and CFB are right angles (perpendicular lines form right angles). AED CFB (All right angles are congruent). ABCD is a parallelogram (A quadrilateral with one pair of sides congruent and parallel is a parallelogram). AD BC (Opposite sides of a parallelogram are parallel). DAE BCF (Parallel lines cut by a transversal form congruent alternate interior angles). DA BC (Opposite sides of a parallelogram are congruent). ADE CBF (AAS). AE CF (CPCTC). REF: geo 1 ANS: Because PROE is a rhombus, PE OE. SEP VEO are congruent vertical angles. EPR EOR because opposite angles of a rhombus are congruent. SPE VOE because of the Angle Subtraction Theorem. SEP VEO because of ASA. SE EV because of CPCTC. REF: b 13 ANS: Parallelogram ABCD, BE CED, DF BFC, CE CF (given). BEC DFC (perpendicular lines form right angles, which are congruent). FCD BCE (reflexive property). BEC DFC (ASA). BC CD (CPCTC). ABCD is a rhombus (a parallelogram with consecutive congruent sides is a rhombus). REF: geo 3
7 14 ANS: Isosceles trapezoid ABCD, CDE DCE, AE DE, and BE CE (given); AD BC (congruent legs of isosceles trapezoid); DEA and CEB are right angles (perpendicular lines form right angles); DEA CEB (all right angles are congruent); CDA DCB (base angles of an isosceles trapezoid are congruent); CDA CDE DCB DCE (subtraction postulate); ADE BCE (AAS); EA EB (CPCTC); EDA ECB AEB is an isosceles triangle (an isosceles triangle has two congruent sides). REF: geo 15 ANS: BD DB (Reflexive Property); ABD CDB (SSS); BDC ABD (CPCTC). REF: ge 16 ANS: Assume parallelogram JMAP with diagonals intersecting at O. Opposite sides of a parallelogram are congruent, so JM AP. JOM and AOP are congruent vertical angles. Because JMAP is a parallelogram, JM AP and since JOA is a transversal, MJO and PAO are alternate interior angles and congruent. Therefore MJO PAO by AAS. Corresponding parts of congruent triangles are congruent. Therefore JO AO and MO PO and the diagonals of a parallelogram bisect each other. REF: 01033b 17 ANS: AB CD, because opposite sides of a rectangle are congruent. AM DM, because of the definition of midpoint. A and D are right angles because a rectangle has four right angles. A D, because all right angles are congruent. ABM DCM, because of SAS. BM CM because of CPCTC. REF: b 4
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