proof geometry

[jen1990]

This is what I have so far:
Statements
1. AC=BC
2. measure of angle ABC=measure of angle CAB

Reasons
1. Given
2. Isosceles Triangle Theorem

I don't know where to go from there.

thank you.

 

[stapel]

Please edit your post, so that the image displays.

Thank you.

Eliz.

 

[stapel]

Okay; you've edited your post so that the image shows up, but it's so tiny that I can't see anything.

It looks like most of your graphic is blank. Try editing your graphic on your computer, deleting all that blank space on the right and at the bottom, and enlarging the actual picture, so we can see it.

That, or provide a detailed description of the image.

Thank you.

Eliz.

 

[jen1990]

On my computer if you click on the image it enlarges. Thanks for trying to help me out.

 

[stapel]

On my computer if you click on the image it enlarges.

Weird.

The image is as follows:

Draw a trapezoid DABE, with DE (the lower base) being wider than AB (the upper base), and the whole thing looking fairly isosceles. Draw line segments AE and DB, with the segments crossing at point C. Erase segment DE.

The question is as follows:

Given: AC = BC
Prove: BE < AE

Eliz.

 

[jacket81]

problem

Okay, we know that CE+BC>BE (I don't know what you call rule, but its when the sum of 2 sides of a triangle are greater then the length of third side)

ANd we know that AE=AC+CE, or, AE-AC=CE by subtracting AC from both sides.
Now since AC=BC, then AE-BC=CE.

Now since we know CE+BC>BE, and AE-BC=CE, substitute in CE, giving you AE-BC+BC>BE, or, AE>BE, we is what we set out to prove!

I hope this helps!

 

[jen1990]

thank you so much!!