Another geometry problem

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matherapist

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Another geometry problem

For an isosceles triangle, ABC, a line is drawn from the vertex, A, that interscects the base BC at a point D where D is not the midpoint of BC.
Prove that AB^2 - AD^2 = BD x CD where AB,AD, BD, and CD are the length of the segments defined by the 2 letters.

Hint: you may have to use an allowed construction at the beginning using only a compass and straightedge.

Posted 151 day ago

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Answers (9)

matherapist

I have no idea what you are talking about. AD is not the height of the isos triangle per the problem statement. If you treat it as such then you have the trivial case of 2 right congruent triangles.

Posted 123 day ago

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atta

An Isosceles triangle, ABC with a height of AD could ensure the existence of the construction irrespective of a compass or straightedge via pure geometrical concepts and theorems. Given two associable right-triangles that are with respect to the height, AD, this can be proven as follows:

AB^2 = BD^2 + AD^2 => AB^2 - AD^2 = BD^2. Likewise, AC^2 = AD^2 + CD^2 => AC^2 - AD^2 = CD^2. In the statement AB^2 - AD^2 = BD * CD, through substitution alone can this statement be proven along with the pythagorean theorem.

J.C

Posted 124 day ago

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matherapist

For proofs, every statement / step needs a reason or other theorem.
The construction/use of the perpendicular to the base of an isos triangle from the vertex angle being the base bisector is a theorem which is proved by congruent rt triangles.
The pythagorean theorem should be stated (though we didn't.
We also used an axiom-Things equal to the same thing are equal.
And I thought you said factoring was your thing in your short bio so I was sort of surprised that it wasn't used. In any case Your thinking and analysis of the problem was on the money.

There are many ways of looking at problems and and in answers perhaps you will see methods you may have overlooked. By the way, it's been about 55 years since I took geometry, but I still remember the concepts and several theorems, and that other theorems i don't recall, exist , and can be looked up.

By the way, that problem was from my HS geometry text

Posted 151 day ago

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math_fanatic

Oh, I see. AE was my construction. You were trying to point out that I did use a construction. I thought I had to use a compass and a ruler to make a construction. As I said, it's been a long time since I knew what all these definitions were.

Posted 151 day ago

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math_fanatic

It is much more efficient though I don't think I would ever have seen the short cut that BE + DE = CD.
I also wouldn't have thought to go straight to the formula for the difference of two squares
right out of the gate.

Did I need to mention that AE was my perpendicular bisector in my proof? It's been a long time since I did geometry so the names of things elude me. I just knew that I could divide the triangle into two identical right triangles with a line down the middle. In this case, I ended up with 3 right triangles because of point D.

Thanks for the tips. I always seem to miss the shortcuts. I hate that :)

Posted 151 day ago

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matherapist

Sampalazza and math fanatic

Sam. You tend to demonstrate by a specific example. These are not really solutions, but are demonstrations.

Math fanatic- You were on the right track and ultimately you got there. Some comments.
You are using the fact that AE in your solution is a perpendicular bisector of the base (construction)
Thus you have your 2 triangles
It is more efficient to consider the following which follows from your first 2 equations.

so AE^2=AD^2-DE^2= AB^2-BE^2 from rt triangles ADE and ABE

Thus AB^2-AD^2=BE^2-DE^2= (BE+DE)(BE-DE)

BE-DE =BD and BE+DE =CD sinc e CE=BE from the construction

Thus The premise AB^2-AD^2=BD*CD is proved

M

Posted 151 day ago

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math_fanatic

I got dyslexic on the last 5 lines of the problem and the site won't let me edit my answer.
They SHOULD read:

4 (AB^2 – AD^2) = BD^2 + 2BDCD + CD^2 – (CD^2 – 2BDCD +BD^2)
4 (AB^2 – AD^2) = BD^2 + 2BDCD + CD^2 – CD^2 + 2BDCD - BD^2

The squared terms on the right side cancel and we’re left with:
4 (AB^2 – AD^2) = 2BDCD + 2BDCD
4 (AB^2 – AD^2) = 4BDCD
AB^2 – AD^2 = BDCD

Posted 151 day ago

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math_fanatic

I solved this problem algebraically without a construction. This is my original solution. I know Matherapist will be able to keep up with the algebra. I apologize to any students who can’t keep up. It’s easy to understand from an algebraic standpoint but it’s a very long and involved process.

Solution:

Assume there is another point E which is the midpoint of BC.
You have formed two right triangles: ADE (with hypotenuse AD) and ABE (with hypotenuse AB).

DE^2 + AE^2 = AD^2
BE^2 + AE^2 = AB^2
AB^2 – AD^2 = BE^2 + AE^2 – DE^2 – AE^2
AB^2 – AD^2 = BE ^2 – DE^2

Now the goal is to get BE and DE in terms of BD and CD.
BE = BC/2
BC = BD + CD
BE = (BD +CD)/2

Isolating DE is a little more involved:
BD + DE + EC = BC
DE = BC – BD – EC

From above, we know that BC = BD + CD
So DE = BD + CD – BD – EC
DE = CD – EC

Now the goal is to get EC in terms of BD and CD.
EC = BC/2 = (BD + CD)/2
So DE = CD – [(BD + CD)/2]
Using a common denominator of 2 we get:
DE = (2CD – BD – CD)/2 = (CD – BD)/2

Now we go back to our original problem with our new values for BE and DE:
AB^2 – AD^2 = [(BD + CD)/2]^2 – [(CD – BD)/2]^2
AB^2 – AD^2 = [(BD + CD)^2]/4 – [(CD – BD)^2]/4
4 (AB^2 – AD^2) = (BD + CD)^2 – (CD – BD)^2
4 (AB^2 – AD^2) = BD^2 + 2BDCD + CD^2 – (CD^2 – 2BCDC +BD^2)
4 (AB^2 – AD^2) = BD^2 + 2BDCD + CD^2 – CD^2 + 2BCDC - BD^2

The squared terms on the right side cancel and we’re left with:
4 (AB^2 – AD^2) = 2BDCD + 2BCDC
4 (AB^2 – AD^2) = 4BDCD
AB^2 – AD^2 = BDCD

Posted 151 day ago

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samzappala

The isosceles I drew has a hypotenuse of 4.59"; the sides are 2.65". The angles are 30, 120, 30. The length of line "ad" is 1.38", cutting angle "a" to 76, 44. The angles on the hypotenuse are 74, and 106. I figured in reference to the smaller division; c/(sin gamma)=2.65/(sin 106)=2.75=1.38/(sin 30)=1.84/(sin 44).
The given formula is 2.65^2-1.38^2=2.75*1.85=7.02-1.9=5.12=5.09. Since there is a difference of .03, and my drawing isn't perfect; I agree.

Posted 151 day ago

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