Geometry Question Bank #2 Splitting Triangles

[nathaliewalton]

Question: In triangle ABC, if BC=3 and AC=4, then what is the length of segment CD? (AC) is drawn down the center of the triangle as a bisector. We know that angle A is 90 degrees and that the bisector creates two 90 degree angles at the bottom

Please help me understand why the answer to the question is 16/3 and not 3.

What I think the explanation provided is trying to say is that even though the triangles are similar, the lengths of segments may not be the same? I somewhat understand this. But in order to get the answer 16/3, You need to compare BC/CA and CA/CD. I do not understand why you set the problem up this way. Please explain. Why would you not just compare BC/CA and CA/CD?

Thanks for your help!

[RonPurewal]

hi - a couple of things about this question.

first, you need to POST AN IMAGE of the question. primarily for the benefit of other posters viewing the problem, but also for us.

if you don't know how to post image files, you can upload them to image hosting websites (such as postimage.org), and then post a direct link to the hosting site here.
directions:
(1) go to postimage.org
(2) click "browse"
(3) find the file on your computer and double click it
(4) click "upload"
(5) when the uploading is complete, copy the "direct link" (which should be displayed on the page, underneath some other codes) and paste it into this thread.

thanks!

Question: In triangle ABC, if BC=3 and AC=4, then what is the length of segment CD? (AC) is drawn down the center of the triangle as a bisector.

are you sure this is an accurate description of the picture?
if the triangle is called ABC, then you can't have "AC" down the middle of it.
or, if "AC" goes down the middle of the triangle, then A and C can't both be corners of the triangle.

But in order to get the answer 16/3, You need to compare BC/CA and CA/CD. I do not understand why you set the problem up this way. Please explain. Why would you not just compare BC/CA and CA/CD?

Thanks for your help!

you wrote BC/CA and CA/CD both times.
it seems that you're trying to offer an alternative option, so it doesn't make sense that you wrote the same two fractions both times. (this may make more sense after you post an image.)

[chip.virnig]

http://www.postimage.org/image.php?v=Pq1Hml5i

[chip.virnig]

I posted the picture of the triangle in question. Anyone have any thoughts? Is there something wrong with the question or is it just worded strangely?

[mxs2009]

There is definitely some information missing from the question as posted. It is impossible for AC to be a bisector of the triangle because ABC is a 3-4-5 right triangle. if AC was a bisector, ABC would be an isosceles triangle. do you mind reposting the question in its entirety?

[chip.virnig]

In triangle ABC to the right, if BC = 3 and AC = 4, then what is the length of segment CD?

http://www.postimage.org/image.php?v=Pq1Hml5i


a) 3
b) 15/4
c) 5
d) 16/3
e) 20/3

After reading the solution, i can now visualize it by rotating the triangle ACD in my head counter clockwise 90 degrees, but this didn't stick out to me at first. I have a decent grasp on the basics around triangle properties, but the similarity between what sides and angles are similar here didn't stick out to me.

[mxs2009]

OK. You can always revert to the Pythagorean Theorem when stuck.

Label segment CD x and Segment AD y.

Now set up two equations:
x^2 + 4^2 = y^2
5^2 + y^2 = (3+x)^2

the two equations simplify to
x^2 - y^2 = -16
-x^2 + y^2 = 6x-16

add the two equations:
0 = 6x - 32
x = 16/3

[RonPurewal]

the solution posted by mxs2009 is good.

alternatively, you can MEMORIZE the following relationships for this special (albeit somewhat rare on this test) situation:

IF THERE'S A LARGE RIGHT TRIANGLE WITH AN ALTITUDE DRAWN FROM THE RIGHT ANGLE TO THE HYPOTENUSE (as there is here):

just memorize the following relationships:
(AB)^2 = BC times BD
(AC)^2 = BC times CD
(AD)^2 = BD times CD [corrected by moderator after original post]

notice that, if you orient the triangle as shown here (big hypotenuse on the bottom), then all three of these are the same:
(thing standing up)^2 = (product of two things in the bottom that share the same endpoint as the thing standing up).

if you know this, then the second one solves this problem in a few seconds:
(AC)^2 = BC x CD
4^2 = 3(CD)
16/3 = CD
done.

this makes for good FLASH CARD material (the general relationships, not anything particular to this problem)

[cyaged]

Ron,

Is BC^2 = BD * CD

supposed to be
AD^2 = BD * CD?

[esledge]

Ron,

Is BC^2 = BD * CD

supposed to be
AD^2 = BD * CD?

Yes, I believe that is what he meant. I'll edit his original message above.

[zdgura]

Maybe it's just me, but those response didn't help at all. Specifically the one that used the pyth theorm. For instance:

Label segment CD x and Segment AD y.

Now set up two equations:
x^2 + 4^2 = y^2
5^2 + y^2 = (3+x)^2

the two equations simplify to
x^2 - y^2 = -16
-x^2 + y^2 = 6x-16


Where's you get the 5^2 from? I'm guessing that it's from the 3,4,5 triangle, but this isn't really explained at all. I guess my question and what I thought was the question of the person who wrote this initial thread was why when you look at this two triangle and you recognize that this is a problem about he relationship between lengths of sides, do you have to set this up as BC/AC = AC/CD. Is there a particular rule to think about how to set these up. I don't want the answer as much as I want to know what drives me to set this problem up this way and not as I initially did which was BC/AC = AC/BC (if I think about it I'm trying to get the relationship between the Base and Height of one triangle to the relationship of the Base and Height of another triangle). To me (and I know that I'm wrong) this is how my set up works. Thoughts?

[tim]

I think everyone who discussed this problem previously just assumed that 5 was so obvious it didn't need to be pointed out. Sounds like you know where it comes from too, so I'm not sure what the problem is..

You should definitely be trying to get the ratio of base:height of one triangle equal to the ratio of base:height of the other. Sounds like you have the right idea here; you just wrote the ratio wrong. Take a closer look and you'll see the approach you are naturally inclined to take ends up giving you the ratio that will eventually solve the problem..

[thiago.ferrari]

Guys,

I'm still stuck with this problem.

I got the Pythagorean Theorem answer, which seems to make sense.

But I don't get one the rational from BC/CA = CA/CD

Let me try explain my rational and see where it is wrong:

If angle ACD is a right angle, then ACB is also a right angle. Additionally, AC is the bisectrix of the angle BAD.

Then:

For triangle ACB the height is AC = 4 and the base is BC = 3
For triangle ACD the height is AC = 4 and the base is CD = ?

Since both are similar, the ratio of base:height should be equal, then BC/AC = CD/AC which BC = CD

So, do you guys know where I got it wrong?

[RonPurewal]

Additionally, AC is the bisectrix of the angle BAD.

this is false.
if this were true, then you would have two 3-4-5 triangles back-to-back.
3-4-5 triangles don't have 45° angles, so, in that case, angle BAD wouldn't be a right angle.

the issue is that you are getting the similarity wrong.
the correspondence is NOT, as you have assumed, between BC in the left-hand triangle and CD in the right-hand triangle (and between AC in both triangles).
instead, BC in the left-hand triangle corresponds to AC in the right-hand triangle, and AC in the left-hand triangle corresponds to CD in the right-hand one.

the problem is that the diagram isn't drawn to scale. if you don't see how these relationships work, then re-draw the triangle yourself so that ABC actually LOOKS like a 3-4-5 triangle. (right now the "3" and "4" look the same, which is the genesis of the whole issue.)
then draw the altitude and this stuff will make more sense.

[joe_mals]

I may be completely missing something but it is my understanding that all corresponding angles must be equal for triangles to be similar. Therefore if I know that two corresponding angles are equal then the two triangles are similar. If AC does not bisect triangle BAD, how would I confirm that triangle ACD and triangle ACB are similar triangles?

In other words, I know triangle acb and triangle acd are right triangles. What other corresponding angle in these two triangles is equal to confirm that these triangles are similar?

What am I missing here? Thanks for your help.