Does Point M Lie on Line BC in This Triangle Bisector Problem?

In summary, the problem involves a triangle ABC with angle A = 60 degrees. Points E and F are the bisectors of angles B and C, respectively. The reflection of point A in the line EF, denoted as M, is to be proved to lie on line BC. The solution involves showing that AFIE, IDCE, and ABDE are all cyclic quadrilaterals, and using the fact that the angle between the normal to EF and the x-axis is 60 degrees.
  • #1
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Homework Statement



Let ABC be a triangle in which angle A = 600. Let BE and CF be the bisectors of the angles B and C, with E on AC and F on AB. Let M be the reflection of A in the line EF. Prove that M lies on BC.

http://img135.imageshack.us/img135/606/sumu.png

The Attempt at a Solution



I didnt really get this problem...i was led to make assumptions and then solve, but that is wrong, i believe.

So can you please help me out?
 
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  • #2


any help?
 
  • #3


I asked this question to one of my friends. This is what he replied-

E is equidistant from AB and BC
F is equidistant from AC and BC
Perpendicular from E to AB (EX), perpendicular from F to AC (FY) and perpendicular from A to EF (AZ) are concurrent.
B+C=120°

I could not infer anything from this. May be you can.
Explain me this if you get it :biggrin:
 
  • #4


Have you got any hint from your teacher?

ehild
 
  • #5


ehild said:
Have you got any hint from your teacher?

ehild

no, this was actually asked in an olympiad...
 
  • #6


Abdul Quadeer said:
I asked this question to one of my friends. This is what he replied-

E is equidistant from AB and BC
F is equidistant from AC and BC
Perpendicular from E to AB (EX), perpendicular from F to AC (FY) and perpendicular from A to EF (AZ) are concurrent.
B+C=120°

I could not infer anything from this. May be you can.
Explain me this if you get it :biggrin:

Erm...
I couldn't get it yet..:frown:
your friend could solve it?
 
  • #7


anyone, please?
 
  • #9


i'm sorry, but i still did not understand...
Of that page i did already know of Incircles and Excircles and relation to the area, Nine Point Circle theorem. But i did not understand how to apply that property...could you please give be a deeper explanation?
 
  • #10


For an equilateral triangle the solution is simple. The incenter coincides with the circumcenter, orthocenter etc and thus EF is parallel to BC with E and F being the mid points of their respective sides. Thus, the reflection of A from EF will lie on BC and will be the mid point of BC. I'm just thinking of how to prove this for a more general case.

One really long, but general method to prove or disprove this would be to assume that the point A lies on the origin. Let the point B lie on the X axis and its coordinates be (x1,0).

As we know the angle A is 60 degrees, the equation of the line AC comes out to be [tex]y=\sqrt{3} x[/tex] and the point C has coordinates [tex]x_2, \sqrt{3} x_2[/tex].

The line BE thus has the equation [tex]y=\frac{1}{\sqrt{3}} x[/tex].

The long winded part is finding the points of intersection of these two line to get the points E and F, thus to get the equation of EF and to find the reflection of A. If this lies on the x-axis then its proven, otherwise not.

There should be a geometrical solution for this which is much simpler, however, the above solution should also work albeit with a considerable amount of work.
 
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  • #11


chaoseverlasting said:
For an equilateral triangle the solution is simple. The incenter coincides with the circumcenter, orthocenter etc and thus EF is parallel to BC with E and F being the mid points of their respective sides. Thus, the reflection of A from EF will lie on BC and will be the mid point of BC. I'm just thinking of how to prove this for a more general case.

I could do that too, but as you the general case isn't working out so well.
 
  • #12


I got it at last. See the picture. Let be β=60-δ and γ=60+δ. The point M is the centre of the incircle. Let be the distance AM=1. Find the angles around M and everywhere in terms of δ, and apply the Sine Law for the yellow and blue triangles to find the lengths AE and AF. Find the coordinates of E and F. Find the tangent of the line EF. Find the angle its normal encloses with the x axis. Find AP and AQ, using the Sine Law again.

ehild
 

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  • #13


ehild said:
I got it at last. See the picture. Let be β=60-δ and γ=60+δ. The point M is the centre of the incircle. Let be the distance AM=1.

ehild

But is doing such an assumption correct?
 
  • #14


Why not? The mirror image stays on the opposite side if you blow up that triangle. Instead of starting from the sides of the triangle, I start with AM. All the other lengths including the sides of the triangle are proportional to it.

ehild
 
  • #15


okay, thank you.
This can be generalized too, like taking AM=x.?
 
  • #16


Of course, you can call it anything. You can express all lengths in terms of this x and delta. All lengths are proportional to this x, which cancels at the end when finding the ratio AQ/AP.

ehild
 
  • #17


ehild said:
I got it at last. See the picture. Let be β=60-δ and γ=60+δ. The point M is the centre of the incircle. Let be the distance AM=1. Find the angles around M and everywhere in terms of δ, and apply the Sine Law for the yellow and blue triangles to find the lengths AE and AF. Find the coordinates of E and F. Find the tangent of the line EF. Find the angle its normal encloses with the x axis. Find AP and AQ, using the Sine Law again.

ehild

There was a mistake in the figure, it is corrected now.

ehild
 

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  • #18


Here is a solution i found in the answer key.

In the figure : http://img692.imageshack.us/img692/5837/sumo.png

AFIE is cyclic. Circle AFC cut BC at D. (BF. BA)=(BI.BE)=(BD. BC). IDCE is
cyclic, angle EIC = angle EDC=60°.

ABDE is cyclic. AF=FD, AE=ED (angle ACF = angle FCD, Ð ABE = Ð EBD). AFDE is kite.
Thus, D is the mirror of A in EF.
 
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  • #19


(BF. BA)=(BI.BE)=(BD. BC)

Which property is this?
 

Related to Does Point M Lie on Line BC in This Triangle Bisector Problem?

1. What is a triangle?

A triangle is a three-sided polygon with three angles. It is one of the basic shapes in geometry.

2. What is the sum of the interior angles of a triangle?

The sum of the interior angles of a triangle is always 180 degrees. This is known as the "Triangle Sum Theorem."

3. How do I find the missing side of a triangle?

To find the missing side of a triangle, you can use the Pythagorean Theorem (a² + b² = c²) if you know the lengths of the other two sides. You can also use trigonometric functions (sine, cosine, tangent) if you know one angle and the length of one side.

4. What is the difference between a right triangle and an acute triangle?

A right triangle has one angle that measures 90 degrees, while an acute triangle has all angles measuring less than 90 degrees. In a right triangle, the side opposite the right angle is called the hypotenuse.

5. How can I determine if three given sides can form a triangle?

To determine if three given sides can form a triangle, use the Triangle Inequality Theorem. It states that the sum of any two sides of a triangle must be greater than the third side. If this is not true, then the lengths cannot form a triangle.

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