Geometry Proof - At it for Hours to No Avail

[meiso]

Homework Statement

Given:

Right triangle ABC with right angle A

segment AD is an altitude of the triangle
segment AE is an angle bisector of angle BAC
segment AF is the median of side BC

note: side BC has points D,E, and F on it in that order

EDIT: I've added a picture of the figure:
http://westchesterccb.com/proofpic.JPG

Prove:

angle DAE is congruent to angle EAF

Homework Equations

N/A

The Attempt at a Solution

Got three similar triangles in ABD ~ DAC ~ DBA

Sorry for the lack of formatting. To anyone who would attempt this proof and possibly solve it, I'd greatly appreciate it.

 

[tiny-tim]

Welcome to PF!

Hi meiso! Welcome to PF!

Hint: that's the same as proving that angle BAD = angle FAC

 

[meiso]

Yes, thank you for that hint. I had realized that and I still have not been able to solve it.
I also have the equations (either from givens or similar triangles and substitution):

angle ABC = angle DAE + angle EAF + angle FAC
angle BAD + angle DAE = angle EAF + angle FAC = 45 degrees
angle BAD = angle BCA

If I could somehow prove angle ABC = angle BAD + angle DAE + angle EAF,
then AF = BF, then BF = FC, and by the base angles theorem,
angle FAC = angle FCA, then by the transitive property, angle BAD = angle FAC, and the rest is easy from there. However, I can't seem to establish that link above.

 

[tiny-tim]

Yes, thank you for that hint. I had realized that and I still have not been able to solve it.

ah … well, that's why it's so important for you to start by telliing us what you have tried

ok, new hint:

you have a right-angled triangle, so draw it in a circle ​

 

[meiso]

Wow. Very elegant tiny-tim. Thank you. BC becomes the diameter of the circle, so BF, FC, and AF are all radii, so their measures are all equal, which is what I needed.

I was helping a high school student whom I tutor with this proof, and I never would have thought of using your method because he has not done anything like that in class. Thank you again, but if there is another method without using a circumscribed circle, I would love to know!

 

[tiny-tim]

… if there is another method without using a circumscribed circle, I would love to know!

ok … alternative hint:

draw FG parallel to AC ​

 

[meiso]

Ok. So FG is drawn with G on segment BA, and intersects two sides of the triangle, so, being parallel to the base, it divides the other two sides proportionally.
Since BF=FC, BF/FC = 1, so BG/GA must equal 1 and and therefore BG=GA.

Following from that, I can use the SAS(BG=GA, two right angles, Reflexive GF) theorem to prove triangle BGF is congruent to triangle AGF. Triangle BGF is similar to triangle BAC (AA Theorem), so angle BFG = angle BCA. Angle AFG = angle FAC because they are alt. int. angles, but angle AFG also equals angle BCA (because angle BCA = angle BFG). By the transitive property (angle BCA = angle BAD already established), then, angle BAD = angle FAC!

This all led from your hint. Please tell me if any of my logic was incorrect (I know it's a lot to sift through) or if there is a shorter way to reach the conclusion.

And thanks again!

 

[tiny-tim]

… Following from that, I can use the SAS(BG=GA, two right angles, Reflexive GF) theorem to prove triangle BGF is congruent to triangle AGF. …

Yes that's ok, but from there on, because I know the circle method, I'd concentrate on sides rather than angles:

FG perp BA, so (SAS) BF = AF, so CF = AF, so angle CAF = ACF