Quadrilaterals Question
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Given that ΔAFB≅ΔEDC, ¯BC || ¯FD, m∠BFE=90°, and points A,F,E are collinear, prove that quadrilateral BCDF is a rectangle.

Statement | Reason |
1.m∠BFE=90° | 1.Given |
2.A,F,E are collinear | 2.Given |
3.m∠AFB+m∠BFE=180° | 3. |
4. | 4.Substitution Property of Equality |
5.m∠AFB=180°-90° | 5.Subtraction Property of Equality |
6.m∠AFB=90° | 6.Algebra (subtract) |
7.ΔAFB≅ΔEDC | 7.Given |
8. | 8.Corr. angles of congruent triangles congruent |
9.m∠AFB=m∠EDC | 9.Definition congruent angles |
10.90°=m∠EDC | 10.Substitution Property of Equality |
11.¯BC || ¯DF | 11.Given |
12.m∠EDC+m∠BCD=180° | 12. |
13.90°+m∠BCD=180° | 13.Substitution Property of Equality |
14.m∠BCD=180°-90° | 14.Subtraction Property of Equality |
15.m∠BCD=90° | 15.Algebra (subtract) |
16.m∠BFE+m∠EDC+m∠BCD+ m∠CBF=360° | 16. |
17.90°+90°+90°+m∠CBF=360° | 17.Substitution Property of Equality |
18.270°+m∠CFB=360° | 18.Algebra (add) |
19.m∠CFB=360°-270° | 19.Subtraction Property of Equality |
20.m∠CFB=90° | 20.Algebra (subtract) |
21.∠BFE, ∠EDC, ∠BCD, ∠CBF are right angles | 21.Definition of right angles |
22.Quad. BCDF is a rectangle | 22.Quad. with 4 right angles is a rectangle |
Grade 10 Quadrilaterals CCSS: HSG-SRT.B.5
- m∠AFB+m∠BFE=m∠AFE
- 90°+90°=180°
- m∠AFB+90°=180°
- 90°+m∠BFE=180°