$AC = BD$

$EC = ED$

$AF = FB$

Angle CAF = 70 deg

Angle DBF = 60 deg

We are looking for angle EFA.

I have found through Geogebra that the required angle is 85 deg. Any ideas how to prove it? I am not so familiar with Geometry :(


Consider the following triangle:-

enter image description here

Let $JM = a, JN = b $ . In this particular $\triangle$, $MK=NL =$ say $x$.

Draw the angle bisector of $\angle J$ , $JO$.

WLOG $a<b$.

Then , by the internal angle bisector theorem , $MR = k_1a , RN = k_1b , KO= k_2(a+x) , OL = k_2(b+x) $.

Obviously , $MN=k_1(a+b) $ and $KL =k_2(a+b+2x)$

Now , locate a point $J'$ along $JL$ , such that $JJ'=\frac{b-a}{2}$ . While this may seem arbitrary , things will become clear soon.

Draw $J'Q$ parallel to $JO$ .


Using similarity in $\triangle $s $JRN$ and $J'SN$ , $SN$ = $\frac{k_1(a+b)}{2}$

This implies that $S$ is the midpoint of $MN$ !

Similarly , we find $QL$ to equal $k_2(\frac{a+b}{2}+x)$ , proving that $Q$ is the midpoint of $KL$ .

Recall that by construction, $J'Q$ is parallel to $JO$.

Thus , we have discovered the fact , that :-

The line joining the midpoints of the opposite sides of a quadrilateral ,when its other sides are equal , is parallel to the angle bisector of the angle formed by extending the other two sides.

Your problem is now trivial .

In your case , $\angle MKL=70 , \angle KLN =60 $

$\therefore \angle KJL = 50 \implies \angle RJN = \angle QJ'N = 25$

External angle $J'QK = 25+60 = \boxed{85} $

  • $\begingroup$ Hope you retained OP's labels of intersection points. $\endgroup$ – Narasimham Apr 19 at 13:02
  • $\begingroup$ @Narasimham No , I'm sorry , I haven't ;) . I can change all the labels , and the image , if you desire, but it would take a while. $\endgroup$ – Sinπ Apr 19 at 13:06
  • $\begingroup$ @Sinπ thank you very much!! No need to change any labels; everything is very clear!! $\endgroup$ – Samuel Apr 19 at 13:38
  • $\begingroup$ Your result was apparently new, I up-voted. In good time you may consider relabelling for an answer that corresponds, esp. it is accepted now :) $\endgroup$ – Narasimham Apr 19 at 13:44

Here' is another solution. Let $\alpha$ and $\beta$ be the angles at $A$ and $B$, resp., let $a$ and $b$ be the length of $AF$ and $AC$, resp. Consider a coordinate system with origin $F$ and let $FB$ the direction of the abscissa. Let $\varphi$ be the angle in question.

Then $$FC=\begin{pmatrix}-a\\0\end{pmatrix}+b\begin{pmatrix} \cos(\alpha)\\ \sin(\alpha)\end{pmatrix}\text{ and } FD=\begin{pmatrix}a\\0\end{pmatrix}+b\begin{pmatrix} -\cos(\beta)\\ \sin(\beta)\end{pmatrix},$$ hence $FE$ the midpoint of $C$ and $D$ is $$\frac12b\begin{pmatrix} \cos(\alpha)-\cos(\beta)\\ \sin(\alpha)+\sin(\beta)\end{pmatrix}.$$ Therefore the slope of $FE$ is $$\tan(180-\varphi)=\frac{\sin(\alpha)+\sin(\beta)}{\cos(\alpha)-\cos(\beta)} =\frac{2\sin\bigl((\alpha+\beta)/2\bigr)\cos(\bigl((\alpha-\beta)2\bigr)}{-2\sin\bigl((\alpha+\beta)/2\bigr)\sin(\bigl((\alpha-\beta)2\bigr)} =-\frac{1}{\tan\bigl((\alpha-\beta)/2)\bigr)},$$ that is $$\tan(\varphi)\cdot\tan\bigl((\alpha-\beta)/2)\bigr)=-1,$$ hence the line $FE$ is perpendicular to one with an angle of $(\alpha-\beta)2$. Thus, $180-\varphi$ and $(\alpha-\beta)/2$ differ by $90$.

In our case $(\alpha-\beta)/2=5$, so the perpendicular line must have an angle of $95$, that is $180-\varphi=95$.

NB: I'm sure there is a simpler way to achieve this general result.


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