Question 817727: How do i prove that triangle BAF and Triangle FCB are congruent given Segment BA is perpendicular to Segment AF, Seg. CF is perpendicular to Seg.BC, and Seg.BA parallel to Seg.CF. I'm confused on how to use perpendicular given to prove.
Answer by LinnW(1048)   (Show Source):
You can put this solution on YOUR website! How do i prove that triangle BAF and Triangle FCB are congruent given Segment BA is perpendicular to Segment AF, Seg. CF is perpendicular to Seg.BC, and Seg.BA parallel to Seg.CF. I'm confused on how to use perpendicular given to prove.
It appears that there is a rectangle BAFC.
Let's see if this is true.
Since BA is perpendicular to segment AF, angle BAF is 90 degrees.
The line that is an extension of AF intersects the two parallel
lines part of the segments BA and CF. This means that angle BAF
equals the corresponding angle near point F. This angle is 90 degrees
so angle CFA is also 90 degrees ( 2 angles on line = 180 ).
Since segment CF is perpendicular to segment BC, BCF is 90 degrees.
We now see we have a four sided figure BAFC with three 90 degree angles.
Since a four sided figure formed by strait lines has a total of 360 degrees,
the fourth angle must be 90 degrees.
So BAFC is a rectangle.
Since BAFC is a rectangle, opposite sides have equal lengths.
BA is congruent to CF, and BC is congruent with AF.
The triangles BAF and FCB have a common side BF.
So all three sides of the triangles are congruent
and the triangles are congruent.
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