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?

kes1968
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

MaxWong
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 :

https://gyazo.com/7bea17ad8e37aa35ee2230ca584f6f47

CPhill
Jun 6, 2017