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Question for sureshbala

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Question for sureshbala

by pakaskwa » Fri Mar 06, 2009 5:17 am
Hi Sureshbala,
There are so many threads to your original posting, so I had to start a new one here.

I just don't think the solution to your question below is correct:

In the figure below, ABC is a right angled triangle. Taking the hypotenuse AC a square is constructed. If O is the center of the square find angle ABO.

Your answer is ABO=45 degree. But if you actually draw the figure on a piece of paper, and move point B, and measure the angle ABO with a ruler, you'll get different numbers. There's no way that there's only one angle.
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by pakaskwa » Fri Mar 06, 2009 5:25 am
Here's another figure posted by 4meonly. I read your explanation on why AB1O = AB2O =AB3O. I'm not convinced.

Simply put, the angle will change when point B is moved along the arc.
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Re: Question for sureshbala

by sanju09 » Fri Mar 06, 2009 6:54 am
pakaskwa wrote:Hi Sureshbala,
There are so many threads to your original posting, so I had to start a new one here.

I just don't think the solution to your question below is correct:

In the figure below, ABC is a right angled triangle. Taking the hypotenuse AC a square is constructed. If O is the center of the square find angle ABO.

Your answer is ABO=45 degree. But if you actually draw the figure on a piece of paper, and move point B, and measure the angle ABO with a ruler, you'll get different numbers. There's no way that there's only one angle.
I have an explanation for this pakaskwa. Since I am not able to show you my drawings right now, so please have a pencil and piece of paper ready for a clear understanding.

Do the following constructions in the figure:

Draw a circle with AC as diameter, and it will pass through B as well, agree!

If you now complete the square by taking AC as one side towards the suggested direction; the sides of the square originating from the points A and C will also be tangent to the circle. With O as the center of square, angle AOC = 90 degrees, this confirms that the drawn circle would also pass through O. If we call AE as the side of square that begins from A, other than AC; then angle ABO = angle OAE (alternate segment theorem); but angle OAE = 45 degrees (AO is along one diagonal of the square). This proves that angle ABO = 45 degrees.
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by pakaskwa » Fri Mar 06, 2009 2:59 pm
sanju09,
Yes I agree it's a circle with AC as its diameter. And the circle circumscribe quadrilateral ABCO.

However, quadrilateral ABCO is NOT necessarily a square, even though there are 2 right angles ABC and AOC.

When B moves along arc AC, angle BAC and BCO will change. Thus angle ABO will change too.

ABO=45 if and only if quadrilateral ABCO is a square.
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by sanju09 » Sat Mar 07, 2009 2:12 am
pakaskwa wrote:sanju09,
Yes I agree it's a circle with AC as its diameter. And the circle circumscribe quadrilateral ABCO.

However, quadrilateral ABCO is NOT necessarily a square, even though there are 2 right angles ABC and AOC.

When B moves along arc AC, angle BAC and BCO will change. Thus angle ABO will change too.

ABO=45 if and only if quadrilateral ABCO is a square.
I will say that you are missing a great deal here, motion of B along the arc AC (not containing O) won't temper the result, a result drawn from the established geometrical theorems. The quadrilateral ABCO is not necessarily a square to prove the said point; as long as angle ABC remains 90 degrees, angle ABO will remain 45 degrees with associated angles adjusting their values to make it happening, under the situations given.
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by pakaskwa » Sat Mar 07, 2009 2:27 pm
I was thinking about this problem last night when I was half into sleep. And I realized where I was wrong. I re-read yours and all the threads in bala's original posting. And I understand why it's contant 45 degree.

Thank you so much sanju09!
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