# Can't understand what I have stumbled upon

#### alevelsprivately

##### New member
Hi!
This is my first post on this website, so lets stick to constructive criticism.
Thanks I was trying to prove Pythagorean Theorem myself, I drew this triangle (just a reference)

Its basically a right angle triangle with a perpendicular angle bisector.

From here what I do is get divide BC into BD = x and DC = y
Now using trig (stupid for this proof but...)
x = ABsin45 = AB/$$\displaystyle \sqrt{2}$$
y = ACsin45 = AC/$$\displaystyle \sqrt{2}$$

so as BC = x+y

$$\displaystyle BC = (AB + AC)/ \sqrt{2}$$

now squaring both sides

$$\displaystyle BC^2 = (AB+AC)^2/2$$

or more generally

$$\displaystyle h^2 = (a + b)^2/2$$

Now out of curiosity I tried this out on the 3-4-5 right angled triangle

and I got

$$\displaystyle \sqrt{(3+4)^2/2} = 4.95$$

which is pretty close to 5
----------------------------------------------------------------------------------
Now my question is,
Can anyone explain me what relationship have I found between these variables or am I just pondering upon a special case
or maybe just a bad way to prove Pythagorean theorem. I am pretty confused as to what this is.
Also, if I have gone wrong somewhere please correct me.

Thanks again for all the help

#### Subhotosh Khan

##### Super Moderator
Staff member
Hi!
This is my first post on this website, so lets stick to constructive criticism.
Thanks I was trying to prove Pythagorean Theorem myself, I drew this triangle

View attachment 13221
(just a reference)

Its basically a right angle triangle with a perpendicular angle bisector.

From here what I do is get divide BC into BD = x and DC = y
Now using trig (stupid for this proof but...)
x = ABsin45 = AB/$$\displaystyle \sqrt{2}$$
y = ACsin45 = AC/$$\displaystyle \sqrt{2}$$

so as BC = x+y

$$\displaystyle BC = (AB + AC)/ \sqrt{2}$$

now squaring both sides

$$\displaystyle BC^2 = (AB+AC)^2/2$$

or more generally

$$\displaystyle h^2 = (a + b)^2/2$$

Now out of curiosity I tried this out on the 3-4-5 right angled triangle

and I got

$$\displaystyle \sqrt{(3+4)^2/2} = 4.95$$

which is pretty close to 5
----------------------------------------------------------------------------------
Now my question is,
Can anyone explain me what relationship have I found between these variables or am I just pondering upon a special case
or maybe just a bad way to prove Pythagorean theorem. I am pretty confused as to what this is.
Also, if I have gone wrong somewhere please correct me.

Thanks again for all the help
You reported:

x = ABsin45 = AB/√2
y = ACsin45 = AC/√2

Which angles in your drawing are 45o and why are those 45o?

#### JeffM

##### Elite Member
You have assumed that the bisector of the right angle in a right triangle meets the hypotenuse in a right angle.

Let's see how that works in a 30/60/90 triangle. Triangle is ABC. The right angle is BAC. I extend the bisector of angle BAC to intersect BC at D.

I now have triangle ABD with angle DAB = 45 degrees, angle ABD = 30 or 60, which entails that angle BDA =

180 - (45 + 30) = 105 or 180 - (45 + 60) = 75, neither of which is 90 degrees.

So you have a fallacy at the very beginning. The only time it is true is in the special case of a right isosceles triangle.

You make it really hard to follow what happens thereafter by never explicitly defining your variables. But I think what you mean is this.

$$\displaystyle h = \text {length of BC.}$$

$$\displaystyle a = \text {length of AB.}$$

$$\displaystyle b = \text {length of AC.}$$

Assuming that is correct, then you say that

$$\displaystyle \dfrac{1}{\sqrt{2}} = sin( \angle DAB) = \dfrac{a}{x} \implies a \dfrac{x}{\sqrt{2}},\ text { and}$$

$$\displaystyle \dfrac{1}{\sqrt{2}} = sin( \angle DAC) = \dfrac{b}{y} \implies b \dfrac{y}{\sqrt{2}}.$$

Now those statements are false unless angle ADB = angle ADC = 90 degrees.

So those statements are wrong except in the special case of a right isosceles triangle. And to prove them correct in that sppcial case, you need the Pythagorean theorem.

And then it gets even goofier.

From the incorrect statements above, you conclude that

$$\displaystyle h = x + y \implies \dfrac{a}{\sqrt{2}} + \dfrac{b}{\sqrt{2}} \implies h = \sqrt{\dfrac{(a + b)^2}{2}},$$

WHICH IS NEITHER CORRECT NOR THE PYTHAGOREAN THEOREM.

EDIT:
Actually, had you worked that last line out, you would have arrived at

$$\displaystyle 2h^2 = a^2 + 2ab + b^2 \implies 2h^2 = a^2 + b^2 + 2ab \implies 2h^2 = h^2 + 2ab \implies h^2 = 2ab.$$

That contradicts the Pythagorean Theorem. So that might have given you a clue that something was off in your reasoning.

Last edited:

Staff member

#### Dr.Peterson

##### Elite Member
I will attach a better diagram

View attachment 13223
Are you aware of the fact that if the angle bisector is perpendicular to the opposite side, the triangle is isosceles?

You can either draw an angle bisector, or an altitude to the hypotenuse; you can't assume the same line is both.

#### Jomo

##### Elite Member
Now draw the diagram (and please make your rt angle look like a rt angle), draw the perpendicular line as you did before and label all the angles. There should only be 3 distinct angles (of which 1 is 90degrees). Now play around with sin2x + cos2x =1 and you will get your proof. Please show it to us! It is very good that you do not just let your teacher give you formulas without wanting to know why it works.

#### JeffM

##### Elite Member
Now draw the diagram (and please make your rt angle look like a rt angle), draw the perpendicular line as you did before and label all the angles. There should only be 3 distinct angles (of which 1 is 90degrees). Now play around with sin2x + cos2x =1 and you will get your proof. Please show it to us! It is very good that you do not just let your teacher give you formulas without wanting to know why it works.
Jomo

I agree that it is excellent that the student is working on his or her own and trying to think in different ways. But unless the student can prove

$$\displaystyle sin^2(\theta) + cos^2(\theta) = 1$$

without relying on the Pythagorean Theorem, the proof is invalid due to circularity. And of course, the proposition is generally false.