Proving Triangle Properties & Calculating Lamp Post Height

In summary: Please clarify.For number 1, I don't understand the clue you gave.For number 2, triangle ABD is right-angled. The length of BD is 30 (half of BC). How does this lead to the length of EC? I think it's because I don't really understand what they mean by the statement "Given that equal [in area] triangles on equal bases have...equal altitudes." Please clarify.If point P is equidistant from AB and BC, does that mean the shortest distance from P to any point on the line AB is equal to the shortest distance from P to any point on the line BC? If I found this distance, how would that lead to congru
  • #1
masterofthewave124
74
0
Hey,

Check out these questions.

1. The point P lies within the angle ABC and is equidistant from AB and BC. Prove that PB bisects the ∠ABC.

If point P is always equidistant from AB and BC, does that mean AB and BC are the same length? If so, that would produce congruent triangles correct?

2. A fence post 4.5 m high casts a shadow of 2.75 m. At the same time a nearby lamp post cast a shadow 13.75 m in length. Find the height of the lamp post.

I need help getting started on this one, I think it has something to do with similar triangles.

Thanks in advance.
 
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  • #2
masterofthewave124 said:
Hey,

Check out these questions.

1. The point P lies within the angle ABC and is equidistant from AB and BC. Prove that PB bisects the ∠ABC.

If point P is always equidistant from AB and BC, does that mean AB and BC are the same length? If so, that would produce congruent triangles correct?
No, it does not. Pick 3 points A, B, C at random and draw ABC, then draw your P and you can see that A and C can lie anywhere along the line AB and BC respectively without changing the condition on P.

You are given that P is equidistant from AB and BC. Do you know what the distance between a point and a line is? Try drawing the line segments those distances refer to.

2. A fence post 4.5 m high casts a shadow of 2.75 m. At the same time a nearby lamp post cast a shadow 13.75 m in length. Find the height of the lamp post.
This is straightforward. Draw it if you can't figure it out.
 
  • #3
For the second question, the shadow that is cast is a diagonal that extends to the base of the triangle? In that case, it's quite an easy similar triangles question.
 
  • #4
masterofthewave124 said:
For the second question, the shadow that is cast is a diagonal that extends to the base of the triangle? In that case, it's quite an easy similar triangles question.
Yes, this is correct. have you worked out this problem?
And also have you worked out the first problem? Do you know what the distance between a point and the line is?
A hint for the first problem: You may need to use congruent triangles to tackle it.
Can you go form here? :)
 
  • #5
VietDao29 said:
Yes, this is correct. have you worked out this problem?
And also have you worked out the first problem? Do you know what the distance between a point and the line is?
A hint for the first problem: You may need to use congruent triangles to tackle it.
Can you go form here? :)

No, I'm fairly confused. If point P is equidistant from AB and BC, does that mean the shortest distance from P to any point on the line AB is equal to the shortest distance from P to any point on the line BC? If I found this distance, how would that lead to congruent triangles?
 
  • #6
OK, I think I may have got it.

Quoted from above,
"If point P is equidistant from AB and BC, does that mean the shortest distance from P to any point on the line AB is equal to the shortest distance from P to any point on the line BC?"

Assuming this is correct, let's call the point from P to the shortest distance on AB, D. And let's call the point from P to the shortest distance on BC, E.

So,

PD = PE (given - equidistant)
PB = PB (shared)
Angle BDP = Angle BEP (shortest distance formed by a line segment from a point to a line is 90 - perpendicular)

Thus, Triangle BDP is congruent to Triangle BEP. (Side-Angle-Side or Hypotenuse-Side theorems).

Therefore, AngleP BD must also equal Angle PBE and so PB bisects Angle ABC.
 
  • #7
I need some help with some more proofs with altitudes.

1. From this image:

http://img124.imageshack.us/img124/1076/altitude18pk.jpg [Broken]

BX and CY are altitudes. Prove that (AB)(AY) = (AC)(AX).

2. From this image:

http://img65.imageshack.us/img65/2723/altitudes27av.png [Broken]

Given that equal [in area] triangles on equal bases have equal altitudes, equal triangles having equal altitudes have equal bases, solve the following for altitude CE.
 
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  • #8
clue for the 1st:
notice that the triangle's size can be calculated in more than one way.

clue for the 2nd:
the triangle ABD. what kind of triangle is it? what can you tell about the length of BD?
 
  • #9
greytomato said:
clue for the 1st:
notice that the triangle's size can be calculated in more than one way.

clue for the 2nd:
the triangle ABD. what kind of triangle is it? what can you tell about the length of BD?

For number 1, I don't understand the clue you gave.

For number 2, triangle ABD is right-angled. The length of BD is 30 (half of BC). How does this lead to the length of EC? I think it's because I don't really understand what they mean by the statement "Given that equal [in area] triangles on equal bases have equal altitudes, equal triangles having equal altitudes have equal bases".

Thanks for your help so far.
 
  • #10
number 1:
take another look at the equation (AB)(AY) = (AC)(AX).
you can also write it like this:
(AB)/(AC) = (AX)/(AY)
does this reminds you of something? (hint: yes)

number 2:
forget about the length of BD... i was going for trigonometric approach...
see the size of the triangle ABC. how many ways can be used to calculate this size?
 
  • #11
greytomato said:
clue for the 1st:
notice that the triangle's size can be calculated in more than one way.

clue for the 2nd:
the triangle ABD. what kind of triangle is it? what can you tell about the length of BD?
greytomato, I should suggest you that you should think twice before giving someone else help. And Please DO NOT confuse the OP... :grumpy:
 
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  • #12
sorry for confuzing you... I'm working on my english skills :)
(size means area)
 
  • #13
masterofthewave124 said:
I need some help with some more proofs with altitudes.

1. From this image:

http://img124.imageshack.us/img124/1076/altitude18pk.jpg [Broken]

BX and CY are altitudes. Prove that (AB)(AY) = (AC)(AX).

There is an easy way and a tough way to do this. Obviously you want the easy way.


Observe triangles BAX and CAY are similar. They share a common angle A and there is a right angle in each, fixing the other angle to be equal. You can now come up with a ratio relating the lengths of corresponding sides which immediately rearranges to give the required relationship.

The hard way involves comparing areas computed two different ways and using Pythagoras' theorem and some algebra. Forget about this unless you want to know for interest's sake.


2. From this image:

http://img65.imageshack.us/img65/2723/altitudes27av.png [Broken]

Given that equal [in area] triangles on equal bases have equal altitudes, equal triangles having equal altitudes have equal bases, solve the following for altitude CE.

The easy way is to use the hint. For triangle ABC, what are two obvious ways to compute the area ? Form an equation and you get CE immediately.

There are other harder ways using similar triangles, Pythagoras theorem and even a trigonometric approach. Don't worry about these, the question clearly wants to guide you to the easy way.
 
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  • #14
Curious3141 said:
There is an easy way and a tough way to do this. Obviously you want the easy way.

The easy way is to use the hint. For triangle ABC, what are two obvious ways to compute the area ? Form an equation and you get CE immediately.

There are other harder ways using similar triangles, Pythagoras theorem and even a trigonometric approach. Don't worry about these, the question clearly wants to guide you to the easy way.

OK, I got number 1. I know what you're trying to hint at for number 2. Just use the basic 1/2bh area of a triangle formula. I guess I'm just not clear on whether AD and CE are actual heights of the triangle. Are all altitudes heights?

Also, can someone confirm the proof I had for the first question in my initial post?

Thanks.
 
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  • #15
Your proof for the first question of your initial post is correct except for your statement that the congruence is of "side angle side" form. It is actually of "side side angle" form which doesn't necessary imply congruence, so go with the "hypotenuse and side" form instead.

Every altitude is a height when the corresponding base is the side that the altitude is perpendicular to.
 
  • #16
Thanks Orthodontist.

I have a couple more problems that I'm stuck on.

http://img235.imageshack.us/img235/1548/triangle1jo.jpg [Broken]

Given that AB ≠ DB. Prove that AB ≠ BC using indirect reasoning.

OK, so I guess I have to start out assuming AB = BC. This leads to saying Angle BAD = Angle BCA (ITT). But, now I have no where to go and I have not yet incorporated the AB ≠ DB in yet. Any insight?

http://img62.imageshack.us/img62/3314/triangle28jk.jpg [Broken]

What's the value of triangle ABD/triangle BGC?

For this one, if I declare a shared base, I don't have a common height.
 
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  • #17
What do the 1 and the 2 mean in the first figure?

In the second figure, in what ways can you divide up the large triangle?
 
  • #18
I don't know what the 1 and 2 mean. That's just how it is from my textbook. My best guess would be it is used to strengthen the fact that Angles BAD and BDA are different, since AB ≠ DB.

For 2, I have tried dividing the triangles up in many ways. If I use AB as my base, triangle ABD is 3x the base of triangle BGC but in no ways are their heights related. Similar result as if I used BC for my base.
 
  • #19
Well, in #1 it is perfectly possible for AB to equal BC even though AB is not equal to BD. There must be additional information, perhaps given by the "1" and "2," that tells you more.

In #2 think broader than that. How can you slice up triangle ABC?
 
  • #20
Hmm, I'll have to ask my teacher if there may have been a typo or simply the 1 and 2 mean something more significant.

Anyways, for #2, instead of slicing up triangle ABC at all, can I not compare each of the triangles ot ABC as a whole and then to each other? So,

triangle ABD = 1/4 triangle ABC (same height, 1/4 base)
triangle BGC = 1/3 triangle ABC (same height, 1/3 base)

Therefore, triangle ABD/triangle BGC = 3/4.

Can someone please verify this method? Thanks.
 
  • #21
masterofthewave124 said:
Hmm, I'll have to ask my teacher if there may have been a typo or simply the 1 and 2 mean something more significant.

Anyways, for #2, instead of slicing up triangle ABC at all, can I not compare each of the triangles ot ABC as a whole and then to each other? So,

triangle ABD = 1/4 triangle ABC (same height, 1/4 base)
triangle BGC = 1/3 triangle ABC (same height, 1/3 base)

Therefore, triangle ABD/triangle BGC = 3/4.

Can someone please verify this method? Thanks.
Looks good, but by the way, we don't divide triangle by another triangle. We, however, can divide the area of a triangle by the area of another triangle. :)
 
  • #22
masterofthewave124 said:
Hmm, I'll have to ask my teacher if there may have been a typo or simply the 1 and 2 mean something more significant.

Anyways, for #2, instead of slicing up triangle ABC at all, can I not compare each of the triangles ot ABC as a whole and then to each other? So,

triangle ABD = 1/4 triangle ABC (same height, 1/4 base)
triangle BGC = 1/3 triangle ABC (same height, 1/3 base)

Therefore, triangle ABD/triangle BGC = 3/4.

Can someone please verify this method? Thanks.
Yes, modulo Viet's comment that is what I meant. You can look at ABC as being composed of three triangles whose area is the area of BGC, or as being composed of four triangles whose area is the area of ABD, so 3 * area(ABD) = 4 * area(BGC).
 
  • #23
Thanks for the verification. I need some confirmation on these questions as well so if you can just take a quick glance at my shortened solutions (the questions are easy for you guys).

1)In ΔABC, the point D divides BC so that BD = DC. A is joined to D and the point E is on AD such that AE:ED = 1:4. I was asked to find the ratio to the following and by way of area of a triangle ratios/proportions, I have my answers beside the assigned.

a. ΔBED: ΔBAE = 4:1
b. ΔBED: ΔCED = 1:1
c. ΔCEA: ΔCAD = 1:5
d. ΔCED: ΔABC = 2:5

2) Determine the value of x and y.

http://img453.imageshack.us/img453/158/unit2test0dp.png [Broken]

Using Triangle Proportion Property Theorem, I came up with the following:

x = 2, y= 20/3

3)
a. Calculate the length of DE.
b. If the area of ΔADE is 50, calculate the area of ΔABC.
c. Determine the area of ΔCEF.
d. Determine the area of parallelogram DEFB.

http://img238.imageshack.us/img238/5162/unit2test6eo.png [Broken]

a. I found that ΔADE is similar to ΔABC primarily by Parallel Line Theorem and found DE to be 4.

b. Consequently, because ΔADE is similar to ΔABC, I found ΔABC's area to be 312.5.

c. I found that ΔADE is similar to ΔCEF primarily by Parallel Line Theorem again and that ΔCEF = 112.5.

d. Simply by subtracting the individual triangle areas from the entire ΔABC, the parallelogram's area is 150. Or through proportion with ΔABD, the same area was found.

It would be a great favour if someone could take a quick look at these. Your help is not going unappreciated!
 
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  • #24
Part b. of 3 is impossible. The maximum possible area of ADE is 4.
 
  • #25
#2 is right
 
  • #26
I have a couple more problems that I'm stuck on.

http://img235.imageshack.us/img235/1548/triangle1jo.jpg [Broken]

Given that AB ≠ DB. Prove that AB ≠ BC using indirect reasoning.

OK, so I guess I have to start out assuming AB = BC. This leads to saying Angle BAD = Angle BCA (ITT). But, now I have no where to go and I have not yet incorporated the AB ≠ DB in yet. Any insight?

Recalling this question, the question is supposed to read: "Given that AB = DB", I'm not even sure how this helps.

With regards to the my last post, yes part b) of the last question is impossible with the values given, but does not affect the method for the succeeding questions, so can someone verify it with the original givens?
 
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  • #27
Again, what do 1, and 2 in your picture stand for? Have you asked your teacher?
Just draw an isosceles triangle ABC, AB = BC. Let D be a point on the line segment AC, [tex]D \not \equiv C, \mbox{and } D \not \equiv A[/tex].
So you'll have: AB ≠ DB BUT AB = BC.
------------
And all the problems you've done above look correct to me. :)
 
  • #28
Thanks for verifying my answers but I don't understand what you're trying to say about the question I am having problems with. My teacher said that the 1 and 2 are negligible but the question did have an typo in that instead of "AB ≠ DB", it's supposed to read "AB = DB". He said it is now possible to prove that AB ≠ BC using indirect reasoning. Did your method account for that?
 
  • #29
OK, so I think I figured it out.

This is original question:

Given: AB = DB. Prove that AB ≠ BC using indirect reasoning.

http://img235.imageshack.us/img235/1548/triangle1jo.jpg [Broken]

Proof (using indirect reasoning):

Start by assuming AB does equal BC.
So, angle BAC = angle BCA.

Then,
DB = BC (since AB = DB),
angle BDC = angle BCD (ITT).

And angle BDA = angle BDC (since angle BAC = angle BDA = angle BDC = angle BCD).

But angle BDA ≠ angle BDC since neither angle is 90 degrees (supplementary angles) - proof that neither is 90 degrees is by the fact that since AB = DB, angle BAC + angle BDA < 180 so each is < 90.

Therefore if angle BDA ≠ angle BDC, then DB ≠ BC and AB ≠ BC.
 
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  • #30
Looks good to me. :)
 

1. What are the basic properties of a triangle?

The basic properties of a triangle include having three sides, three angles, and the sum of the interior angles being equal to 180 degrees. Additionally, triangles can be classified based on their side lengths and angle measurements.

2. How do you prove that a triangle is a right triangle?

A triangle can be proven to be a right triangle if it follows the Pythagorean theorem, which states that the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. This can be checked by measuring the side lengths and using a calculator to square and add them.

3. What is the formula for calculating the height of a lamp post using triangle properties?

The formula for calculating the height of a lamp post using triangle properties is h = (d * tanθ) + a, where h is the height of the lamp post, d is the distance from the base of the lamp post to the point where the angle is measured, θ is the angle of elevation from the ground to the top of the lamp post, and a is the height of the observer's eye level.

4. Can triangle properties be used to calculate the height of any object?

Yes, triangle properties can be used to calculate the height of any object as long as the distance from the base of the object to the point where the angle is measured and the angle of elevation are known. This method is commonly used in surveying and engineering.

5. How can we use triangle properties to determine if two triangles are congruent?

Two triangles are congruent if they have the same three side lengths or the same three angle measurements. This can be determined using the Side-Side-Side (SSS), Side-Angle-Side (SAS), or Angle-Side-Angle (ASA) congruence criteria. If these criteria are met, then the triangles are congruent.

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