Triangles Question
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In the following figure, ¯AB || ¯CE, ∠BFE and ∠CDE are supplementary, and points A,F,E are collinear. Also, ∠BCD≅∠CDE and ¯AF≅¯ED. If BC=2CD and CD=2ED, prove that BD⋅BF=AB⋅BC (¯BD is not drawn).

Statement | Reason |
1.∠BCD≅∠CDE | 1.Given |
2.BC=2CD | 2.Given |
3.BCCD=2 | 3.Division Property of Equality |
4.CD=2ED | 4.Given |
5.CDED=2 | 5.Division Property of Equality |
6.BCCD=CDED | 6.Transitive Property of Equality |
7. | 7.SAS Similarity Theorem |
8.BDCE=BCCD | 8.Ratios corr. sides of similar triangles are equal |
9.¯AF≅¯ED | 9.Given |
10.¯AB || ¯CE | 10.Given |
11.∠BAF≅∠CED | 11. |
12.∠BFE and ∠CDE are supplementary | 12.Given |
13.m∠BFE+m∠CDE=180° | 13. |
14.m∠BFE=180°-m∠CDE | 14.Subtraction Property of Equality |
15.A,F,E are collinear | 15.Given |
16.m∠AFB+m∠BFE=180° | 16. |
17.m∠BFE=180°-m∠AFB | 17.Subtraction Property of Equality |
18.180°-m∠CDE=180°-m∠ABF | 18.Transitive Property of Equality |
19.-m∠CDE=-m∠AFB | 19.Subtraction Property of Equality |
20.m∠CDE=m∠AFB | 20.Multiplication Property of Equality |
21.∠CDE≅∠AFB | 21.Definition of Congruent Angles |
22.ΔAFB≅ΔEDC | 22. |
23.¯AB≅¯CE | 23.Corr. sides of congruent triangles congruent |
24.AB=CE | 24.Definition of Congruent Segments |
25.¯BF≅¯CD | 25.Corr. sides of congruent triangles congruent |
26.BF=CD | 26.Definition of Congruent Segments |
27.BDCE=BCCD | 27.Earlier result |
28.BDAB=BCBF | 28.Substitution Property of Equality |
29.BD⋅BF=AB⋅BC | 29.Multiplication Property of Equality |
Grade 10 Triangles CCSS: HSG-SRT.B.5
- ΔBCD ~ ΔCDE
- ΔBCD ~ ΔCED
- ΔBCD ~ ΔDEC
- ΔBCD ~ ΔDCE