Similar Triangles I was reading a book and came across the following question...
$\text{If in triangles ABC and DEF, we have } \angle A =\angle D \ \text{and AB:DE=BC:EF, then prove that } \angle C= \angle F\ \text{or} \ \angle C+\angle F=180^\circ $
I don't have any specific ideas to prove this ... I have so far only shown that if both are acute angled triangles then first condition holds ......By taking three cases 
$(i)\angle E = \angle B$,  $(ii)\angle E > \angle B$,  $(iii)\angle E < \angle B$
It's easy to prove the first
but I don't know how to prove this more generally for all possible configurations
....
Any hints would be appreciated..
 A: AB : DE = BC : EF implies AB : BC = DE : EF. This further means $\dfrac { \sin C }{\sin A} = \dfrac {\sin F}{\sin D}$ by sine law.
A = D implies $\sin C = \sin F$
Then, either $C = F$ or C = $180^0 – F$
A: Remember.  If you have a point $E$,  a line $DF$, and a length $k$ then either i) there are two points on $DF$ that are $k$ distance for $E$.  ii) The is exactly one point on $DF$ that is $k$ distance from $E$ and the segment formed is perpendicular to $DF$.  or iii) There are no points of $DF$ that are $k$ distance from $E$.
So lets try to construct triangle $DEF$ from what we know.  Plot point $D$.  Construct a line $L$ through $D$.  Construct $\angle D = \angle A$.  Plot a point $E$.  Find the ratio of $AB$ to $DE$.  Calculate the distance $k$ so that $k$ is the proportional to $BC$ as $AB$ is to $DE$. 
Now try to figure which points on $L$ that are $k$ distance from $E$ can be candidates for point $F$.  There must be at least one as we can make a triangle similar to $ABC$.  (So iii) is out of the question.).  If there is one such point then call it $F$ and $\angle F = \angle C$ is a right angle as is $ABC$ and $DEF$ and $ABC$ are similar AND $\angle F + \angle C = 180$.  
The third option is there are two points $F$ and $F'$.  Wolog $D, F', F$ are the colinear order.  
Call $\angle DFE = \angle F$ and $\angle DF'E = \angle F'$.
As $EF = EF'$ then $EFF'$ is isoceles.  So $\angle EF'F = 180 - \angle F' = \angle F$.
Note one of the triangles $DEF$ or $DEF'$ is similar to $ABC$ so either $\angle C$ is equal to either $\angle F$ or to $\angle F' = 180 - \angle F$
(see image)

In the image:  $\angle F = \angle EF'F = 180 - \angle DF'E $.
These are the only two possible triangles where $\angle D = \angle A$, $DE = k*AB$ and $DF$ (or $DF'$) $= k*BC$.  Now one of these two triangles must be the triangle that is similar to $\triangle ABC$.
So either $\angle C = \angle F$ or $\angle C = DF'E = 180 - \angle F$.
A: Another attempt:
$ \triangle ABC$:   
Extend $AB$ to $E$,  and extend $AC$ to $F$.
Consider  $ \triangle ABC$, and $\triangle AEF$.
Given: $AE:AB= EF:BC $.
Then   $EF = BC(AE/AB)$.
Draw a line parallel to $BC$ through $E$, let it intersect line $AF$ in $F'$.
Intercept Theorem (Similar triangles):
$EF' : BC = AE: AB$; 
Hence $EF' = BC(AE/AB)$. 
We conclude : length $EF$ equals length $EF'$.
1) Length $EF$ > distance from $E$ to line $AB$ extended.
A circle with center $E$ and radius $r: =$ length $EF$ intersects line $AB$ extended in two distinct  points $F$  and $F'$ .
A) $ \triangle ABC$ is similar to $\triangle  AEF'$.
$(EF' || BC)$. Corresponding angles are equal.
B)$\triangle EFF'$ is isosceles , $r$ = length $EF'$ = length $EF$, the two base angles at $FF'$ are equal.
$\angle EFA = 180° - \angle EF'A$.
2)  Distance from  $E$ to $AB$ extended equals  $r$.
Circle touches, i.e. $F = F'$, only one solution : 
$\angle EFA = \angle EF'A = 90°$.
3)  $ r$ < distance, no solution.
Comments welcome.
A: Superimpose $∆ABC$ on $∆DEF$. As $\angle A = \angle D$, (AB,DE) and (AC,DF) overlap. Since the question involves 'lengths' and condition is imposed on C and F, we have to find the possible positions in which C and F can be placed to satisfy the condition. Let us assume that the condition ($AB:DE = BC:EF$) is true for certain points C and F. Taking E as the centre and EF as the radius draw an arc. Name the points as $F_1$ and $F_2$ . Similarly for C we will have $C_1$ and $C_2$. $∆EF_1F_2$ and $∆BC_1C_2$ are isosceles triangle. Corresponding to each value of C there exists two values of F. Since there exists two 'nominees' in each case and we know that one of them is a similar triangle so either $\angle BC_2A = \angle EF_2D$ or $\angle BC_1A = \angle EF_1D$.
Case 1.
$$\angle BC_2A = \angle EF_2D$$
$$\angle EF_2F_1 = 180° - \angle EF_2D = 180° - \angle BC_2A$$
Therefore either $$\angle F = \angle C$$ or $$\angle F + \angle C = 180°$$
Case 2.
$$\angle BC_1A = \angle EF_1D$$
$$\angle EF_2F_1 = \angle BC_1A$$
$$\angle EF_2D = 180° - \angle EF_2F_1 = 180° - \angle BC_1A$$
Therefore either $$\angle F = \angle C$$ or $$\angle F + \angle C = 180°$$
When $\angle BCA = 90°$ , $\angle F = 90°$.
