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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?

Jun 6, 2017

#1
+1

Are these spurious 6s supposed to be angle symbols, viz., should it read:

If ∠DAE = ∠DFE, what is ....

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Jun 6, 2017
edited by Badinage  Jun 6, 2017
edited by Badinage  Jun 6, 2017
#2
+2

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

Jun 6, 2017
#3
+2

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 :   Jun 6, 2017
edited by CPhill  Jun 6, 2017
edited by CPhill  Jun 6, 2017
#4
+1

Here is CPhill's picture:  Jun 6, 2017
edited by hectictar  Jun 6, 2017