# Triangle question

1. Jun 6, 2012

### praharmitra

Hey guys,

This is NOT homework. I remember solving this question many years ago (at least 10 years ago). I am trying to recall the solution again and am just not able to. The question is -

In a triangle ABC, AD is the angle bisector of angle BAC. AB = CD. Prove that angle BAC = 72 degrees.

A diagram is attached.

NOTE - I am actually not even sure the question has enough information as given. I am just recalling from memory. If you think more information is required give an appropriate reason as to why you think that is true.

Usually, the way I see if the question has enough information is that I try to construct a triangle given the above properties. If I can construct a unique triangle, then of course the information is enough. Else not.

Anyway, give it a try.

https://www.dropbox.com/s/wpvei01hw1nr25l/2012-06-06%2000.39.06.jpg

2. Jun 6, 2012

### HallsofIvy

Staff Emeritus
Looks like non-sense to me. Do this: draw a right triangle, ABC, with right angle at A. At vertex C, use a protractor with radius set at length AB to mark point D on BC. D will be between B and C because the length of a leg of right triangle is always less than the length of the hypotenuse. You now have exactly the situation shown in your picture but the angle is 90 degrees, not 72. Clearly that can be done taking any acute angle at A.

3. Jun 6, 2012

### praharmitra

Well, but with that construction AD will not be the angle bisector of angle A.

4. Jun 6, 2012

### DaveC426913

72 degrees is 1/5 of a circle, or the complementary angle of a pentagram. My guess is that the solution will involve a pentagram or star of some sort.

5. Jun 6, 2012

### coolul007

this sounds very much like golden ratio

6. Jun 6, 2012

### praharmitra

Hi guys,

I figured out the problem with the question. There is not enough information to solve this problem. You can see this by doing the following construction. Draw the line AD first (This can be any length, for this argument atleast). Now draw the two equal angles BAD and DAC on either side of AD. Again, the angle could take any value. Extend the sides AB and AC such that BDC forms a straight line and BD = DC.

Now, in this triangle, AB > CD. Now rotate the side BDC around the point D such that the length of CD increases and that of AB decreases. We can always do this. Since the rotation is continuous, we will always have some angle of rotation at which AB = CD.

We then have constructed exactly the triangle described above! Since the angle is arbitrary, the question is incomplete.

I now recall what the original question was. There was an additional given -

angle ABC = 2 X angle BCA

Try and solve it now!