Hello there! I've been having a bit of trouble with my proof problems lately. I was wondering if someone could explain these few to me just so I could get the hang of these problems? I can send proof that I did some on my own in terms of proofs just to prove I'm not cheating because I know many people do that here.
Here are the problems I'm requesting and a screenshot of the diagram:
1. Suppose that the carpenter measured and found that AB = DC and AD = BC. Can you
conclude from this that the wall is rectangular? Explain.
2. Suppose that the diagonals were measured instead, and that AC was found to equal BD.
Can you conclude from this that the wall is rectangular? Explain.
Suppose that the wall is measured and that AB = DC, AD = BC, and AC = BD. Mark the figure
as needed to answer the following:
Why is ∆ABC – ∆DCB?
Why is angle ABC = angle DCB?
Why is ABCD a parallelogram?
Why is angle ABC = angle ADC and angle DCB = angle DAB?
Why is angle ADC = angle DAB?
Why is ABCD a rectangle?
I'm really, really sorry that there's more than one. I'm just really confused on this topic and this website is my last resort...
Thank you so much for your time and help <3
+2407 Sure, here's a hint for the first one. :))
The defination of a rectangle is that all 4 angles are 90 degrees. If oposite sides are the same, does that guarentee the angles to be 90 degrees? Try looking at other 4 sided shapes. (Rhombus, square, parallelograms, kite)
=^._.^=
Thanks! That helped, but I'm still having trouble with the other problems. Can anyone spell them out for me? :) in a good way, of course
+526 In △ABC and △DCB,
∵ AB = DC, AC = BD and BC = BC
\({AB\over DC}={AC\over BD}={BC\over BC}=1\)
∴ By SSS (side-side-side) rule of similarity
△ABC ~ △DCB
If △ABC ~ △DCB then they're also congruent.
As a result by cpct (corresponding-part-of-congruent-triangle) we can prove that
∠ABC = ∠DCB
Since AB‖CD and BC‖AD and opposite sides are also equal therefore, ABCD can be classified as a parallelogram.
As ABCD is a parallelogram the opposite angles are equal... hence ∠ABC = ∠ADC and ∠DCB = ∠DAB
I hope I made it clear enough.
You're welcome :)
