# Quadrilaterals with equal sides $$AC = BD$$

$$EC = ED$$

$$AF = FB$$

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 :(

## 2 Answers

Consider the following triangle:- Let $$JM = a, JN = b$$ . In this particular $$\triangle$$, $$MK=NL =$$ say $$x$$.

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

WLOG $$a.

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 $$JO$$ .

$$J'N=JN-JJ'=\frac{a+b}{2}$$.

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

• Hope you retained OP's labels of intersection points. – Narasimham Apr 19 at 13:02
• @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. – Sinπ Apr 19 at 13:06
• @Sinπ thank you very much!! No need to change any labels; everything is very clear!! – Samuel Apr 19 at 13:38
• 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 :) – 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.