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The problem is as follows:

The figure from below shows a triangle $ABC$. Find the angle $x$ on $\angle BCA$.

Sketch of the problem

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&20^{\circ}\\ 2.&12^{\circ}\\ 3.&15^{\circ}\\ 4.&16^{\circ}\\ 5.&18^{\circ}\\ \end{array}$

This problem seems to be approached forming an isosceles by an auxiliary construction or something by drawing a line from $A$ to a point $P$ in $BC$. But I dont know if that's the right approach for this. Can someone help me exactly how to relate it with the sum of the sides given in the triangle?.

Please an answer which would help me the most is some which is not relying on trigonometry. Can someone help me with a drawing to solve this thing?.

Chris Steinbeck Bell
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1 Answers1

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Step 1:

Construct point $E$ on $\overline{AC}$ s.t. $\overline{AB}=\overline{BE}$. Therefore $\angle AEB=2x\Longrightarrow \angle EBC=x\Longrightarrow\overline{BE}=\overline{EC}$.

Step 2:

Construct $\triangle EFC$ ($\triangle EFC\cong\triangle ADB$) as shown in the picture. We then know that $B, F, E$ are on the same line because

  1. $\angle DEB=\angle FEC$
  2. $D,E,C$ are also on the same line

Step 3:

Note that $\overline{AD}=\overline{EF}$ and $\overline{AB}=\overline{BE}$. Therefore $\overline{BE}+\overline{EF}=\overline{BF}=\overline{BC}$. At this point we know that $\triangle BFC$ is an isosceles with base angle $4x$.

Hence we have $9x=180^{\circ}\Longrightarrow x=20^{\circ}$.

enter image description here

  • Howdy!. Thanks you're my savior. I didn't know you're from TW. I love Taipei, I longed to visit and study in NTU since my major is chemistry. The construction which you have made below the triangle is very clever, I didn't imagined such can be done. But I wonder, can this be solved by using similarity or congruence?. – Chris Steinbeck Bell Oct 25 '20 at 23:32
  • I am not sure about this. At least I am not able to come up with a solution using them. –  Oct 26 '20 at 02:54