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The problem is as follows:

$AD$ bisect angle $\angle BAC$. Point $E$ which lies on $BC$ is generated by the $DE$ which is a bisector of $\angle ADC$. Using this information and the figure from below, find the angle which is made by $DE$ and $BC$.

Sketch of the problem

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&100^{\circ}\\ 2.&90^{\circ}\\ 3.&110^{\circ}\\ 4.&120^{\circ}\\ \end{array}$

I'm totally lost in this problem. What exactly should be the way to approach this?.

The problem not only lies in solving but how can it be solved relying only in euclidean geometry postulates?. The thing here from what I can spot is that there might come into play notable angle in triangles. Such as the angle formed by an exterior bisector or an interior bisector.

Can someone help me in a way that I can understand without much fuss?. Please include a drawing in the answer because I'm lost here.

Suspiciously $\angle ABC = 2 \angle BCA$. Is this an implicit hint which makes us to conclude to draw a segment and build an isosceles there?. I don't know. Does it needs congruence or what?. Please guide me, I need help step by step.

Chris Steinbeck Bell
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2 Answers2

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Hint:

  • The internal angles of any triangle sum to $180^\circ$.
  • The angles on a straight line sum to $180^\circ$.

That's all you need.

lonza leggiera
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0

You can just chase angles.

Find $\angle ECD$ (and $\angle ACD$)
Find $\angle BAC$ from $\triangle ABC$
$AD$ is a bisector, so half this for $\angle CAD$
Find $\angle ADC = 2\alpha$ from $\triangle ACD$
Calculate $\alpha$
Find $x$ from $\triangle ECD$

Joffan
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