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I was given steps to construct a figure:

1.) Construct a horizontal ray AB and a segment AC at an angle to the ray. Locate point D anywhere on ray AB and construct the segment CD.

2.) Construct the angular bisectors of angle ADC and ACD.

3.) Construct the intersection of E of the two bisectors.

4.) Construct the perpendicular of ray AB to point E.

5.) Construct the intersection of F of the perpendicular and AB. Construct the segment EF and then erase the two angular bisectors.

6.) Construct a circle with center E passing through F (this circle should appear tangent to both rays and CD)

The point E is moving along a certain entity, which in fact is the bisector of angle BAC. Prove that this is indeed so, i.e. that the point E lies on the bisector of angle BAC.

I constructed the figure, I want to make sure it is correct.

Also, I'm not entirely sure how to prove that E lies on the bisector of angle BAC. Any ideas would be appreciated!

Constructed Figure

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  • $\begingroup$ Consider two of the three touching points. The line from these to the midpoint of the inner circle must be orthogonal to the corresponding sites of the triangle. So, the distance from $E$ to both sides must coincide. So, $E$ must lie on the bisector of the angle between the sides. $\endgroup$ – Peter Apr 19 '16 at 20:09
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You are essentially asking for a proof that the three angle bisectors of a triangle are concurrent. The proof is easy.

The key idea is that a point lies on the bisector of $\angle D$ iff it is equidistant from the lines $AD,CD$. So by construction the point $E$ is equidistant from the lines $AD,CD$ and also equidistant from the lines $AC,CD$.

Hence it is also equidistant from the lines $AC,AD$. Hence it lies on the bisector of $\angle CAD$, otherwise known as $\angle BAC$.

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