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#1 Re: Help Me ! » Geometry Proof » 2015-08-12 10:22:43

Sure, contents list would be helpful!

Quick last question, if ACH, BHD and HAB are congruent (SAA) how is it that AHB = 90?

#2 Re: Help Me ! » Geometry Proof » 2015-08-11 10:55:09

Thanks, but i don't really understand how you got AB = HD = CH or how you got ACP + BDP = 90.

#3 Help Me ! » Geometry Proof » 2015-08-10 09:35:25

Replies: 39

Two circles are externally tangent at point P. Segment CPD is parallel to common external tangent AB. Prove that the distance between the midpoints of AB and CD, is AB/2.

two circles

only without the thin drawn in lines. I couldn't find a better photo.

#4 Re: Help Me ! » Triangle Proofs » 2015-08-05 12:14:10

Is r=(1+sqrt3)/2? And is Cos 36 = 1/2+1/2*r, Cos 72=1/(2+2r)

I checked it, and it didn't seem to work out. Where is it that I went wrong?

#5 Re: Help Me ! » Triangle Proofs » 2015-08-04 11:08:42

Erm... I'm kinda stuck. I got to part (c) and well I got the equation r^2+xr-1=0. Even using the quadratic or completing the square, i get stuck with a square root with a lot of variables.

I want to check to make sure the equation is right, and what i got for r is also right. Then I need a bit of help with the rest.

Super thanks!!

#6 Help Me ! » Triangle Proofs » 2015-08-02 04:27:01

Replies: 21

In triangle ABC, angle A = 36 degrees and angle B = angle C = 72 degrees. Let BD be the angle bisector of angle ABC.

(a) Prove that BC = BD = AD.

(b) Let x = BC and let y = CD. Using similar triangles ABC and BCD, write an equation relating x and y.

(c) Write the equation from Part (b) in terms of r=y/x and find r.

(d) Compute cos 36 degrees and cos 72 degrees using Parts a-c. (Do not use a calculator!)

I got part (a), you can easily find and use the missing angles to prove BC = BD = AD. But I need a bit of help with the other parts.


#7 Re: Help Me ! » Trig Functions » 2015-07-29 01:42:00

Thank you so much!

(i think I figured out the third... still working on it though!)

#8 Help Me ! » Trig Functions » 2015-07-28 02:38:30

Replies: 3

Why does cot^2(x) + 1 = csc^2(x) for any x such that x is not an integer multiple of 180 degrees, where cot=cotangent and csc=cosecant?

Cotangent=(cos(x))/(sin(x)) and Cosecant=1/(sin(x))

My understanding is (cos^2(x))/(sin^2(x))+1=1/(sin^2(x)), multiplying the equation by sin^2x to get;

cos^2(x)+sin^2(x)=sin^2(x), but that doesn't get me anywhere with why that's true for all x that's not a multiple of 180. A little help please?


#9 Re: Help Me ! » Triangles » 2015-07-23 08:34:33

Those are great proofs. But is there a way using coordinate geometry? For example putting triangle ABC in the coordinate plane so that A is the origin and one of the equal sides aligns with the x-axis.

I tried to go on from there, but I'm not sure what to do.

#10 Re: Help Me ! » some squares and a triangle... » 2015-07-08 17:03:06

bob bundy wrote:

Imagine rotating triangle TAC through 90 degrees, anticlockwise, around point A.

T rotates to B and C rotates to V, so one triangle becomes the other.

So TC must also rotate 90 to become BV, hence they are perpendicular.


How do we know that you have to rotate it 90 degrees?

#11 Re: Help Me ! » 3d geometry? » 2015-06-30 15:05:01

I used a simpler algebra way to find x. Because you could use Pythagorean Theorem and write the base as 16+2x. Then since we also have the base as 20-2x. Setting them equal to each other we get x=1. Then use the ratios y/(y+10)=x(10-x) to find y.

#13 Re: Help Me ! » 3d plane in a cube » 2015-06-22 09:04:08

Quick question, if plane through C, P, and Q intersect AD at S, and the same plane through C, P, and Q intersects DH at R, which two dimensional figures do they make?

#14 Re: Help Me ! » spacial geometry » 2015-06-22 08:10:22

I just chose a name. _sheepishly_ I'm not actually a champion at math, but I love math, so I guess I get some credit.

#15 Re: Help Me ! » spacial geometry » 2015-06-21 07:04:25

bob bundy wrote:

To get the ratio of volumes you cube the length SF.

I'm also working on this problem.

Where did the "S" come from?

#16 Re: Help Me ! » Tangent Triangle Proofs » 2015-06-09 10:41:46

Is there a way to see if someone else has posted the same problem, so there are not multiple problems, that are the same, being posted?

#17 Help Me ! » Tangent Triangle Proofs » 2015-06-08 12:04:21

Replies: 7

Let the incircle of triangle

be tangent to sides
, and
, and
, respectively. Prove that triangle
is acute.

Tangent, I'm new to this forum, I don't know whether the diagram came out.

I got a hint saying to compute the angles of triangle

in terms of the angles of triangle
, but after doing so, got me nowhere.

If I could get some help, that would be great!

#18 Re: Help Me ! » 4 secants circle geometry » 2015-06-01 10:11:14

@bob bundy, I was wondering how you got angle DCB = 180 - (x + y) => BCQ = x + y.

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