1. ## Help With Geometry

The following link will take you to a file that has a screenshot of my problem. I'm not allowed to use trig.

2. ## Re: Help With Geometry

First label the vertex at the 'x' $F$.
$\left\{ \begin{array}{l}m(\angle AEB=40^o)\\m(\angle ADB=50^o)\\m(\angle DFE=70^o)\end{array} \right.$

Now leave it to you the explain those values and finish the question.

3. ## Re: Help With Geometry

Thank you so much for your help, but I don't know where to go from there.

5. ## Re: Help With Geometry

I'm not sure why you would "label the vertex at the 'x' F" when it is already labeled "E". Oh, I see. DenisB is not referring to the "x degrees" that you are to determine, he is referring to the point where the two line segments, AE and BD, intersect.

I would use the fact that the measures of the angles in a triangle sum to 180 degrees repeatedly. You are given that angle BAE has measure 60 degrees and angle ABD has measure 50, angle AFB has measure 180- 60- 50= 70 degrees. By the 'vertical angle theorem', angle DFE is also 70 degrees. We have that angle ABE has measure 50+ 30= 80 degrees so angle 180- 60- 80= 40 degrees. There are many more triangles- keep doing that.

6. ## Re: Help With Geometry

I did that, and this is where I got. https://docs.google.com/a/tsc.k12.in...it?usp=sharing

7. ## Re: Help With Geometry

Lol I'm kinda bad at geometry cause I started a month and a half ago but heres my steps
1. Find Triangle CEA, we know that two of the angles are 20° each so that leaves us with the last angle being 140° we'll call this angle <E+X

2. Find Triangle CDB, we have two angles which are 20° and 30°, 180-50 is 130° (the angle that is 130 is <CDE and <CDB)

3. Let's call angle EDB, "Y"

4. When we compare Triangle CEA and Triangle CDB, we want to focus on the third sides, so 130° and 140°
CEA is comprised of X + __ = 140
CDB is comprised of Y + __ = 130

5. Then I realized that you can do substitution, because the difference between 140 and 130 is ten, Y= X-10

6. Let us focus on the center triangle (the one with X in it) it's Y + X + 70 = 180
Substitute Y for X-10,
X-10+X+70= 180 -> 2x-10 = 110 -> 2x=120 -> X=60

7. I know I'm late and you probably already discussed this question in your class already :/

8. ## Re: Help With Geometry

Originally Posted by Zammyz
4. When we compare Triangle CEA and Triangle CDB, we want to focus on the third sides, so 130° and 140°
CEA is comprised of X + __ = 140
CDB is comprised of Y + __ = 130

5. Then I realized that you can do substitution, because the difference between 140 and 130 is ten, Y= X-10
You can't do that Zammy: it assumes x = y !

Try again. Hint: x = 30. See if you can get that...

Also, capital letters are used to represents points,
NOT values; hence my use of x and y, NOT X and Y.

9. ## Re: Help With Geometry

Can you write down the steps that you did cause I initially went on the forums to look at other people's solutions and work to get better

10. ## Re: Help With Geometry

Originally Posted by Zammyz
Can you write down the steps that you did cause I initially went on the forums to look at other people's solutions and work to get better
First label $\overline{DB}\cap\overline{AE}=\{F\}$.
You tell us why each of the following is true.
$\left\{ \begin{array}{l}m(\angle AEB=40^o)\\m(\angle ADB=50^o)\\m(\angle DFE=70^o)\end{array} \right.$

If you cannot the explain why you cannot.

11. ## Re: Help With Geometry

AEB = 40° because 60+30++50= 140, 180-140 = 40
ADB = 50° because 60+20+50= 130, 180-130 = 50
DFE = 70° because 60+50 = 110, <F is 70° and by vertical angles it also = 70

What other reasoning's did you use?

The only other way I've seen to get 30 is get triangle BDC, assume <C is 20, 180-50 = 130 and subtract 50 to get 80, now angle D in triangle DEF is 80+70+X = 180

12. ## Re: Help With Geometry

Consider the bisector AH of angle EAB in triangle EAB, H on EB

Show AH is parallel to DE using CD/DA = CE/EH

To show CD/DA = CE/EH, consider point G on AB extended where BG = BC

Triangles GCA and EAB are similar