Question 1183493: ............A/\...............
............../.\.................
............./...\................
............/.....\...............
.........../.......\..............
........../.........\.............
.....D./...........\.F..........
------------>---------------
......./...............\..........
....../.................\.........
...B.------->------.C.......
Given : |DF is parralel to |BC
Prove : AD = AF
-- --
DB FC
Statements
1. |DF is parallel to |BC
2. Angle ADF is congruent to Angle B
3. Angle A is congruent to Angle A
4. Triangle ADF is similar to Triangle ABC
5. AC AB
..-- = --
..AF AD
6. AF+CF AD+DB
..----- = -----
..AF AD
7. FC DB
..+1-- = -- +1
..AF AD
8. FC DB
..-- = --
..AF AD
9. AD*AF FC AD*AF DB
..----- * -- = ----- * --
..DB*FC AF DB*FC AD
10. AD AF
..-- = --
..DB FC
Reasons
1. Given
2. Corresponding Angles Postulate
3. ___________
4. ___________
5. Definition of Similar Polygons
6. Segment Addition Postulate, Substitution
7. Simplification
8. ___________
9. Multiplication Property of Equality
10. Simplification
Answer by Solver92311(821)   (Show Source):
You can put this solution on YOUR website!
This is complete nonsense. With only the figure and the one given fact, you cannot arrive at the conclusion required.
Consider the following two figures:
In Figure 1, Triangle ABC is isosceles because I drew it that way, but in Figure 2, A'B'C' is not isosceles. However, both figures satisfy your given conditions, namely a triangle with a line segment parallel to the triangle's base. Yet in one case, AD is indeed equal to AF, but in the other case A'D' is most assuredly not equal to A'F'.
Therefore, in general, your desired conclusion is not true.
John
My calculator said it, I believe it, that settles it
From
I > Ø
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