In acute-angled triangle ABC, let D be the foot of the altitude from A, and E be the midpoint of BC. Let F be the midpoint of AC. Suppose 6 BAE = 40◦ . If 6 DAE = 6 DF E, what is the magnitude of 6 ADF in degrees?
Because D and F are the mid-points of BC and AC respectively, DF || AB. Therefore, angle AEF should equal 40 deg.
Let G be the intersection point of AD and EF.
Angle DAE + 40 deg = angle AGF
Angle DFE + angle ADF = angle AGF
Because angle DAE = angle DFE,
angle ADF should equal 40 deg :D
Max.....E is the mid-point of AB, not D....so..... CE/EB = CF/FA ....
Which would mean that EF ll AB.....instead of DF ll AB .....
Then angle FED = angle ABE and angle GDE = angle ADB
So by AA congruency
Triangle DGE is similar to triangle DAB
Thus....angle DGE = angle DAB
And angle DAB = angle DAE + angle EAB
So.....this means that
Angle DGE = angle DAE + angle EAB
But, by the exterior angle theorem, angle DGE also = angle DFE + angle GDF
And angle DAE = angle DFE
So.....this means that
angle DAE + angle EAB = angle DFE + angle GDF
Subtracting like angles on either side of the equation, we have that
angle EAB = angle GDF
But angle EAB = 40°....so angle GDF has the same measure
But angle GDF = angle ADF
And Max's assertion that angle ADF = 40° is correct, after all !!!
**Thanks, Max, for some of the hints you gave.....you actually did 90% of the proof....I just "cheated" off you a little.....LOL!!!!!!
Here is a "quasi" pic :
https://gyazo.com/7bea17ad8e37aa35ee2230ca584f6f47