Geometry problem (1 Viewer)

[iStudent]

This is probably actually 2U, but since I thought of this while doing a 4u question, I placed it here
(feel free to move it if required)

Suppose for any triangle, you find a point such that x = y (refer to below diagram)
would alpha = gamma? (i.e. would a line drawn from it it bisect angle A)

And what about the converse? (i.e. the bisector of an angle bisects the line). That is, if alpha = gamma would x = y?

If it's true, is there a proof for this?
(and if not, why not? if applicable)

 

[aDimitri]

definitely not. pretty easy to see why, just draw an exaggerated case. place point D up near A and look at line BD. the section the subtends angle gamma is much much much shorter than the section that subtends angle alpha.

 

[iStudent]

Sorry I worded that poorly. D (the intersection between the lines) is supposed to be the mid point of BC. (so it lies on BC)
Can you rephrase because I don't understand what you mean by "up near A"

 

[aDimitri]

D isn't there at all. i mean construct line BD so that D lies on AC, but D is close to the point A.

 

[D94]

You can easily design a case where that doesn't hold true.

For example, if AC = CD, and angle ACD is 90 degrees, then angle ADC = 45, and hence angle DAC = 45 degrees (which is gamma). But if you claim alpha = gamma, then angle BAC = 90, and therefore we don't have a triangle any more. So clearly alpha cannot equal gamma.

NB: D is the point that intersections BC.

 

[iStudent]

Then D would have to be the mid point of AC. (so you can't make it closer). I'm still not sure what you mean. :s

 

[aDimitri]

i drew a diagram to show what i mean by constructing BD

 

[aDimitri]

it's in my previous post

 

[iStudent]

I see thanks.

 

[iStudent]

D isn't there at all. i mean construct line BD so that D lies on AC, but D is close to the point A.

But one of the conditions is that D is the midpoint

 

[iStudent]

How about a different situation

Lines AB and AC intersect each other. The bisector of acute angle BAC cuts the interval BC at D. Does BD = CD?

(so you have a triangle this time)

 

[aDimitri]

no it doesn't, that's exactly what my diagram showed...

 

[iStudent]

no it doesn't, that's exactly what my diagram showed...

In your diagram BD doesn't bisect angle ABC though. Nor is D a midpoint

 

[aDimitri]

well change D to E, and make the bisecting line intersect at D. It clearly doesn't bisect line BE

 

[D94]

How about a different situation

Lines AB and AC intersect each other. The bisector of acute angle BAC cuts the interval BC at D. Does BD = CD?

(so you have a triangle this time)

No, it doesn't.

aDimitri's diagram actually illustrates your scenario. But I'll explain it another way. Let's say you have an isosceles triangle ABC, where as you have described, BAC is the acute angle where we can call that the apex of the isosceles triangle. Also, it's obvious that the other two angles are equal (since isosceles). Now, draw a line that bisects angle BAC straight down to meet BC (the base) and that obviously bisects BC, which you called point D. Now, move point B up the side AB (which is one of the two equal sides of the isosceles triangle) and you can see that as you move it up or down, BD does not equal CD.

The attached image will illustrate that.

 

[iStudent]

Ah I get it. Thanks
I must have trouble understanding english, haha.