How do I show that there is a mental ellipse at an angular point of a triangular area which touches the opposite side at its middle point and bisects the adjacent sides?

In Short:

Here's a shorter way to show that there is a mental ellipse at an angular point of a triangular area which touches the opposite side at its middle point and bisects the adjacent sides:

1. Draw a triangle ABC with vertex A at the angular point and with AB and AC as the adjacent sides.

2. Draw the perpendicular bisectors of AB and AC, which intersect at point O.

3. Draw a line through point A that is perpendicular to line AO, and let it intersect BC at point D.

4. Draw a line through point A that is perpendicular to line AD, and let it intersect line AO at point E.

5. Draw a circle with center O and radius OA.

6. Draw a tangent to the circle at point E.

7. Let F be the point where the tangent intersects line BC.

8. Show that the distance from point F to the two foci of the ellipse is constant for all positions of point A.

9. Conclude that the locus of points F, as A moves along the angular point of the triangle, is an ellipse.

10. Therefore, there is a mental ellipse at the angular point of the triangle which touches the opposite side at its middle point and bisects the adjacent sides.

In Long:

To show that there is a mental ellipse at an angular point of a triangular area which touches the opposite side at its middle point and bisects the adjacent sides, we can use the following steps:

 1. Draw a triangle ABC with vertex A at the angular point and with AB and AC as the adjacent sides.

2. Draw the perpendicular bisectors of AB and AC, which intersect at point O.

3. Draw a line through point A that is perpendicular to line AO, and let it intersect BC at point D.

4. Draw a line through point A that is perpendicular to line AD, and let it intersect line AO at point E.

5. Draw a circle with center O and radius OA.

6. Draw a tangent to the circle at point E.

7. Let F be the point where the tangent intersects line BC.

We need to prove that the locus of points F, as A moves along the angular point of the triangle, is an ellipse.

To prove this, we need to show that the distance from point F to the two foci of the ellipse is constant for all positions of point A.

Let G be the midpoint of BC, and let H be the intersection of line AO with the circumcircle of triangle ABC.

Then, we have:

- OE = OA cos(AOE) = OA cos(90°-AOD) = OA sin(AOD) = OG sin(2A)

- HD = AD cos(AHD) = AD cos(90°-AED) = AD sin(AED) = AG sin(A)

- OH = OA cos(AOH) = OA cos(90°-ABC) = OA sin(ABC) = 2OG sin(A)

Now, we can express the distance from point F to the two foci of the ellipse in terms of these lengths:

- Distance from F to focus at O: OF = OF1 = OE + EF = OE + AE tan(AOE) = OE + AO tan(AOD) = OG sin(2A) + OA tan(AOD)

- Distance from F to focus at H: HF = HF1 = HD + DF = AD sin(A) + AF sin(AHD) = AG sin(A) + AE sin(AED) = AG sin(A) + OA sin(ABC)/cos(AOE)

We can simplify these expressions using the fact that AG = OG, and substituting for OE, HD, and OH:

- OF = OG sin(2A) + OA tan(AOD) = OA sin(2A)/(cos(AOE)cos(AOD)) + OA tan(AOD)

- HF = OG sin(A) + OA sin(ABC)/cos(AOE) = OA (sin(A)cos(ABC) + sin(ABC)cos(AOE))/(cos(AOE)cos(AOD))

We can simplify these expressions further by using the fact that AOD = ABC/2, and substituting for cos(ABC) and cos(AOE):

- OF = OA [sin(2A)/cos(AOD) + tan(AOD)] = OA 2sin(A)/cos(ABC)

- HF = OA [sin(A)/cos(AOD) + sin(ABC)/cos(AOE)] = OA (sin(A) + sin(ABC))/(cos(AOD)cos(AOE))

Now, we can see that both OF and HF are expressed in terms of OA, sin(A), and sin(ABC), which are all constants for a given triangle. Therefore, the distance from point F to the two foci of the ellipse is constant for all positions of point A, which means that the locus of points F is an ellipse.

Hence, we have shown that there is a mental ellipse at an angular point of a triangular area which touches the opposite side at its middle point and bisects the adjacent sides.