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I'm working on this plane geometry problem:

$ \bigtriangleup AEF$ is a triangle, $\angle A = 60^\circ$, $\angle F = 40^\circ$, $\angle E = 80^\circ$, points $K$ and $C$ are on line $AF$, $\angle CEF=20^\circ$, $\angle KEF=40^\circ$, $O$ is the mid-point of $AK$, prove that the length of $OC$ is half of $EF$.

I can prove this by using trigonometric formulas. My question is: is it possible to prove it without using the knowledge of trigonometric formulas?

Here is a figure I draw using Geogebra.

enter image description here

TShiong
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Xiaoyong Guo
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2 Answers2

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Let $\overline{EF}=1$ for simplicity. Also note that $\triangle AEC$ is equilateral.

Let $H$ on $EF$ such that $\overline{EH} = \overline{EC}$.

enter image description here

Internal Bisector Theorem on the isosceles triangle $\triangle EKF$ gives

$$\overline{KF} = \frac{\overline{KC}}{\overline{KF}-\overline{KC}}.\tag{1}\label{2}$$

Similarity $\triangle CHF \sim \triangle EKF$ leads to

$$\overline{KF}=\frac{1-\overline{AC}}{\overline{KF}-\overline{KC}}.\tag{2}\label{3}$$

Equating \eqref{2} and \eqref{3} gives $$\overline{KC} = 1-\overline{AC},\tag{3}\label{4}$$ which is equivalent to your thesis. In fact we have \begin{eqnarray} \overline{OC} &=& \frac{\overline{AC}-\overline{KC}}2+\overline{KC}=\\ &=&\frac{\overline{AC}+\overline{KC}}2 =\\ &\stackrel{\eqref{4}}{=}&\frac12. \end{eqnarray}

dfnu
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    Thank you very much for your solution. I think the key is KC=HF. Still it's hard for me to come up with this solution. Also I don't remember this internal bisector theorem, I did some google search, it is quite useful. – Xiaoyong Guo Jan 04 '23 at 03:11
  • Can I accept both solutions? – Xiaoyong Guo Jan 04 '23 at 03:16
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    I choose to accept the second solution since it uses less knowledge. But still thank you so much for you solution, which is very helpful and inspiring, it taught me about the interior bisector theorem. – Xiaoyong Guo Jan 04 '23 at 03:19
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enter image description here

Let $K'$ be a point on $AE$ such that $KK' \perp AE$.

Since $\angle AK'K = 90^\circ$, $AK' = AO$.

As $\triangle AK'O$ and $\triangle AEC$ are equilateral triangles, $OC = K'E$.



Let $K''$ be a point on $EF$ such that $KK'' \perp EF$.

Since $\triangle KEK' \cong \triangle KEK'' \cong \triangle KFK''$, $OC = K'E = \frac{EF}{2}$

$\therefore$ The length of $OC$ is half of $EF$.

Anonymous
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