A property of right triangles

[johnnyluvsmath]

Hi, there’s this question I’ve been stuck on for a while now. I started it but I’m not sure where to go from there. Here is the question and here’s what I did, although I’m sure it’s completely wrong. I would appreciate it if someone could give me a few clues so I can solve the problem.

 

[Steven G]

The measure of a length is a number, nothing special--just a number.

(2*3)(4*5) = 2*4*2*5*3*4*3*5 is so wrong! You never used similar triangles. You never used the triangles! You started off with the formula you wanted to prove. You should have just stopped there.

(2*3)(4*5) = (6)(20)=120 while 2*4*2*5*3*4*3*5 = 8*2*5*3*4*3*5 = 16*5*3*4*3*5 = 80*3*4*3*5 = 240*4*3*5= 960*3*5= 2880*5 = 14400 which is much more than 120.


You do not distribute when you have a product!

 

[Steven G]

You asked for clues.
1) Do not use the equation you want to prove.
2) Use the given diagram
3) Find congruent triangles.
4) Divide corresponding sides and say wow that is what I was asked to show.

 

[Dr.Peterson]

Hi, there’s this question I’ve been stuck on for a while now. I started it but I’m not sure where to go from there. Here is the question and here’s what I did, although I’m sure it’s completely wrong. I would appreciate it if someone could give me a few clues so I can solve the problem.

The problem says to prove this using properties of vectors; I don't see that either you or Jomo have taken any steps in that direction.

The first thing I need to know is, how much have you learned about vectors? What operations do you know, and what properties? That will help us see how to advise you.

 

[pka]

The problem says to prove this using properties of vectors; I don't see that either you or Jomo have taken any steps in that direction. The first thing I need to know is, how much have you learned about vectors? What operations do you know, and what properties? That will help us see how to advise you.

TO: johnnyluvsmath, please read the above adnomination.
This one of the most important theorems: \(\dfrac{m(\overline{AC})}{m(\overline{DC})}=\dfrac{m(\overline{DC})}{m(\overline{BC})}\).
You can use \(\vec{AD}+\vec{DC}=\vec{AC}\) & \(\vec{DC}\cdot\vec{DC}=(m(\overline{DC}))^2\)

 

[johnnyluvsmath]

The problem says to prove this using properties of vectors; I don't see that either you or Jomo have taken any steps in that direction.

The first thing I need to know is, how much have you learned about vectors? What operations do you know, and what properties? That will help us see how to advise you.

I learnt addition and scalar multiplication, associative, commutative, distributive.

 

[Dr.Peterson]

I learnt addition and scalar multiplication, associative, commutative, distributive.

So, no dot product or anything like that? I'd think something like that would be necessary (as in pka's suggestion), to deal with perpendicularity.

 

[johnnyluvsmath]

So, no dot product or anything like that? I'd think something like that would be necessary (as in pka's suggestion), to deal with perpendicularity.

Ah yes I learnt that as well, I forgot to include it.

 

[johnnyluvsmath]

Ah yes I learnt that as well, I forgot to include it.

i learnt cross product as well

 

[Dr.Peterson]

Ah yes I learnt that as well, I forgot to include it.

Okay, now try using it. Start with pka's suggestion in post #5, and if that doesn't work, try some other dot product. Make sure you include some vectors whose dot products will be zero.

This is not going to be immediately obvious, so you just have to be willing to try things without knowing what will happen.

 

[johnnyluvsmath]

Okay, now try using it. Start with pka's suggestion in post #5, and if that doesn't work, try some other dot product. Make sure you include some vectors whose dot products will be zero.

This is not going to be immediately obvious, so you just have to be willing to try things without knowing what will happen.

So I tried some stuff and understand the theorem itself but I can't figure out where to start and how to incorporate vector properties into proving this theorem

 

[Dr.Peterson]

So I tried some stuff and understand the theorem itself but I can't figure out where to start and how to incorporate vector properties into proving this theorem

You haven't learned yet how to get the most out of this site. If you show the stuff you tried, we'll be able to see whether you are going off in a totally wrong direction, or just making some little mistake while doing useful things.

But here's what I said last:

Start with pka's suggestion in post #5, and if that doesn't work, try some other dot product. Make sure you include some vectors whose dot products will be zero.

What I did was first to give easier names to the vectors: [MATH]\vec{a} = \overrightarrow{AC}[/MATH], [MATH]\vec{b} = \overrightarrow{AD}[/MATH], [MATH]\vec{c} = \overrightarrow{AB}[/MATH], [MATH]\vec{d} = \overrightarrow{DC}=\vec{a}-\vec{b}[/MATH].

Then, express the goal in terms of these: [MATH]|\vec{a}|^2 = |\vec{b}||\vec{c}[/MATH]|; and keep in mind that since b and c are collinear, [MATH]\vec{b}\cdot\vec{c}= |\vec{b}||\vec{c}|[/MATH].

Then use these vectors to state which angles are right angles, which is what I suggested focusing on. Then aim toward the goal.

 

[pka]

You need to have a diagram handy to which to refer. I will not use vectors.
Angles \(\angle A~\&~\angle B\) are complementary; as are angles \(\angle A~\&~\angle ACD\).
Thus we get similar triangles \(\Delta ACB \approx \Delta ADC\approx \Delta CDB\).
Can you use proportional parts to finish the question.

 

[Dr.Peterson]

You need to have a diagram handy to which to refer. I will not use vectors.
Angles \(\angle A~\&~\angle B\) are complementary; as are angles \(\angle A~\&~\angle ACD\).
Thus we get similar triangles \(\Delta ACB \approx \Delta ADC\approx \Delta CDB\).
Can you use proportional parts to finish the question.

The problem, of course, says these two things:





I, too, would never bother with vectors for this problem unless I were forced to. But it is interesting once you've done it.

 

[johnnyluvsmath]

You haven't learned yet how to get the most out of this site. If you show the stuff you tried, we'll be able to see whether you are going off in a totally wrong direction, or just making some little mistake while doing useful things.

But here's what I said last:

What I did was first to give easier names to the vectors: [MATH]\vec{a} = \overrightarrow{AC}[/MATH], [MATH]\vec{b} = \overrightarrow{AD}[/MATH], [MATH]\vec{c} = \overrightarrow{AB}[/MATH], [MATH]\vec{d} = \overrightarrow{DC}=\vec{a}-\vec{b}[/MATH].

Then, express the goal in terms of these: [MATH]|\vec{a}|^2 = |\vec{b}||\vec{c}[/MATH]|; and keep in mind that since b and c are collinear, [MATH]\vec{b}\cdot\vec{c}= |\vec{b}||\vec{c}|[/MATH].

Then use these vectors to state which angles are right angles, which is what I suggested focusing on. Then aim toward the goal.

I tried your suggestion and I think I was able to advance though I think im blocked again. Here’s what I did, please let me know if it makes sense and how I could advance further.

 

[Dr.Peterson]

I tried your suggestion and I think I was able to advance though I think im blocked again. Here’s what I did, please let me know if it makes sense and how I could advance further.

You made a couple mistakes at the start. It is not true that [MATH]\vec{a}\cdot\vec{b}= 0[/MATH] or [MATH]\vec{a}\cdot\vec{d}= 0[/MATH]. You should have one zero dot product for each right angle you listed; those pairs of vectors are not perpendicular.

I didn't name your e and f, but expressed everything in terms of a, b, and c, since those are what we want in the conclusion; and I used only one pair of vectors for each right angle. So I expressed your e as a - c, and didn't use your f.

So I think that's my key suggestion: express all vectors in terms of only a, b, and c, and express the two right angles as two dot products, in addition to expressing the collinearity of b and c as an equation. Then solve for |a|^2.

 

[johnnyluvsmath]

You made a couple mistakes at the start. It is not true that [MATH]\vec{a}\cdot\vec{b}= 0[/MATH] or [MATH]\vec{a}\cdot\vec{d}= 0[/MATH]. You should have one zero dot product for each right angle you listed; those pairs of vectors are not perpendicular.

I didn't name your e and f, but expressed everything in terms of a, b, and c, since those are what we want in the conclusion; and I used only one pair of vectors for each right angle. So I expressed your e as a - c, and didn't use your f.

So I think that's my key suggestion: express all vectors in terms of only a, b, and c, and express the two right angles as two dot products, in addition to expressing the collinearity of b and c as an equation. Then solve for |a|^2.

i tried your suggestion but i’m still stuck. no matter what I do it doesn’t give me the answer im looking for

 

[Dr.Peterson]

Did you try distributing these?


These will give you some very useful equations. One will have a clear connection to your goal. Then you can manipulate the other side, aiming toward the goal, [MATH]|\vec{a}|^2 = |\vec{b}||\vec{c}| = \vec{b}\cdot\vec{c}[/MATH].

By the way, the bulk of your work is based on taking two quantities you know are zero, and setting them equal. That is generally not a good idea, because it actually loses information, namely that they are not merely equal, but equal to zero. So I would avoid that.

 

[johnnyluvsmath]

Did you try distributing these?
View attachment 24843View attachment 24844

These will give you some very useful equations. One will have a clear connection to your goal. Then you can manipulate the other side, aiming toward the goal, [MATH]|\vec{a}|^2 = |\vec{b}||\vec{c}| = \vec{b}\cdot\vec{c}[/MATH].

By the way, the bulk of your work is based on taking two quantities you know are zero, and setting them equal. That is generally not a good idea, because it actually loses information, namely that they are not merely equal, but equal to zero. So I would avoid that.

I did it! i think, so I arrived at the answer i was looking for but I have a slight doubt in how I arrived to the answer.

 

[Dr.Peterson]

Beautiful! That looks very much like what I wrote out.

Wasn't that fun ... eventually?