How to find the length of a segment in a quadrilateral made by two triangles?

Question

The problem is as follows:

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\textrm{3 cm}\\ 2.&\textrm{4 cm}\\ 3.&\textrm{5 cm}\\ 4.&\textrm{6 cm}\\ \end{array}$

I'm stuck at trying to find the proper relationships in this figure based on the triangle congruence. Can someone help me here?

The thing is that I'm stuck on how to find $BE$?.

The only relationships which I could spot were:

$AC=AD$

and triangle $BCE$ is isosceles. But nothing more than that?. What else can be done here?.

Please include a drawing in your answer I don't know exactly how to relate a triangle congruence here, which seems to be the intended method of solution?. Can this be approached relying only in euclidean geometry?.

Answer 1

The trick about this problem is that the given angles and lengths show $ABCD$ is inscribed in a circle. This is because $$AD=DE \Longrightarrow \widehat{DAC}=\widehat{DEA}$$ $$\widehat{DEA} = \widehat{CEB}$$ $$BC=CE \Longrightarrow \widehat{CEB}=\widehat{CBD}$$ $$\Longrightarrow \widehat{DAC} = \widehat{CBD}$$ This is enough to show that the four points $A,B,C \& D$ are on the same circle.

What follows is interesting. From the given figure: $$\widehat{BCA} = \widehat{ACD} =\alpha$$ And now that we know the points are on the same circle, we can see that $$ \widehat{ABD} = \widehat{ACD} \quad \& \quad \widehat{ADB} = \widehat{ACB}$$ So: $$ \widehat{ABD} = \widehat{ADB} \Longrightarrow AD=AB$$ Now, the problem tells us $AB=6$ and $AD=DE$ , so we have $DE=6$, and subtracting from the given length of $BD$ we see that $BE=5$.