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The problem is as follows:

The figure from below represents a quadrilateral $ABCE$. Using the information, $BE=BC$, $\angle\,BAC=\angle\,ADB=60^{\circ}$ and $DC=10\,m\,AE=10\,m$. Find the angle $x$.

Sketch of the problem

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&125^{\circ}\\ 2.&140^{\circ}\\ 3.&130^{\circ}\\ 4.&120^{\circ}\\ \end{array}$

I couldn't really find a relationship other than spot an isosceles triangle on $\triangle BEC$. This would mean that $\angle BEC= \angle BCE$. I also could spot that the $\triangle ABD$ is an equilateral. But other than that I'm stuck. What sort of identity or construction here is required to solve this problem?.

Can somone help me?. Can this be solved relying only in Euclidean postulates?. Please an answer must include a drawing because in this figure I can't find exactly where to look for, but I think that it is related with congruence or maybe similarity.

Chris Steinbeck Bell
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2 Answers2

3

Since $\angle BAD = \angle BDA = 60^\circ$, $\triangle ABD$ is equilateral, as you have observed.

We have $AB = BD$, $AE = DC$, $BE = BC$.

By SSS, $\triangle ABE \cong \triangle DBC$.

Now $\angle BAE = \angle BDC$. What is $\angle BDC$?

player3236
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1

The following proportionately drawn figure shows how the problem has been set up. ( Labels as in given problem ):

Two equilateral triangles whose sides are scaled to a ratio

$$\dfrac{BE}{BA}=\dfrac{BC}{BD}$$

are rotated about fulcrum vertex $B$ from position $E$ to $C$ through $60^{\circ}$.

The yellow triangle can be considered to be rigid hence congruent (like when made of plastic or cardboard).

So the external $\angle BAE $ in given figure is $120^{\circ} $ that gets carried forward. Option 4.

enter image description here

Narasimham
  • 40,495