Revision as of 02:58, 17 September 2009

HW 3

Does anybody know what the ratio in #1 is supposed to come out to be? I've tried everything and can't get it to work. - Mark

Mark, check page 38 of the notes. The problem is actually Theorem 28.

Does anyone have any hints for #2. I can't seem to figure it out, I may be misunderstanding some of the directions. -Mary

I figured out how to do #2. Add the distances from the legs of the angle to the point p, and subtract that number by the distance from point p to leg inside the angle. -Craig

Do you mean number 2 the geometer's sketchpad? Or do you mean number two as in the proof section?? -Brittany

I know a lot of people have been having trouble with this problem. I know someone was going to email Uli last night. I would just continually check on here for help. - Dana

I'm not sure if you are having trouble drawing the figure or figuring out the equation. So I'll try to cover both, for drawing the figure start with two points and then make a segment between them. Then choose one of the points and the segment and ask for a circle with radius. Do the same thing again only using the other point, then the third point is the intersection of the circles. As far as the formula is concerned, use addition and subtraction. Hope this is helpful.--Shore 14:20, 16 September 2009 (UTC)

Does anyone have something for problem 6. I'm having trouble with it for some reason. -Jon

Hey Jon, number 6 is exactly like the proof in the book for case one...the congruent triangles are simply on the outside of triangle(ABC). Just connect AX,BX,PX,and CX and look at congruent triangles formed! --Jrhaynie 15:01, 16 September 2009 (UTC)

For number 8, what two pairs of similar triangles do you use to prove part b? - Jon Maser

Jon, for part b i used that triangle AGF is similar to EHF and that triangle BGF is similar to DHF. Hope that helps! -Jennie

I am struggling with number 5. I know I should draw a triangle DEF that is similar to ABC, but I'm not sure how to go about proving any of the angles are equal just based on the proportion of the sides. Any hints? -Lauren

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