Thread: Isosceles trapezoid problem

Originally Posted by mjoshua

That is one explanation yes, however you said that the triangles mentioned above were congruent and they clearly weren't.

Yes, the ones I mentioned are congruent.
How could they not be?

Additionaly, that would mean that ACD is 25, how can this be???

Yes, ACD is equal to BDC which is 25. They are mirror-images.

Base angles of an isosc trapezoid are supposed to be congruent the one base angle BCD would be 60 and if DBC is 95 that would mean ADC is 120 which is not equal to 60.

The base angles are.... BCD=ADC and DAB=CBA

For BCD to equal ADC, ACD needs to be 85 not 25.

No, ADC=25+35=60 degrees=BCD

So why is this happening?

If you draw a vertical line through the point of intersection of the diagonals, perpendicular to both of the sides DC and AB
and noting that the left-hand side of the trapezoid is a mirror-image of the right-hand side, then

Angle BCA=angle ADB
Angle ADC=angle BCD
Angle BDC=angle ACD and so on....

They are not a mirror image--if DBC is 95 then BDA is 95 since alt. int. angles congruent (this is NOT mirrored!! BCA is 35!!)

Originally Posted by mjoshua

They are not a mirror image--if DBC is 95 then BDA is 95 since alt. int. angles congruent (this is NOT mirrored!! BCA is 35!!)

The post title said "isosceles trapezoid".

Yes it did and does, so.... why are the base angles not working out to be equal???

I have 95 + 25 and 35 + 25 these arent equal??

Originally Posted by mjoshua

Yes it did and does, so.... why are the base angles not working out to be equal???

I have 95 + 25 and 35 + 25 these arent equal??

I already told you what the base angles were.

Over and out

You know, I really wish instead of just saying "this is the answer" someone could show me why, step-by-step, everyone else gets a fully worked out solution to their problems and I'm not even for that, I'm asking for someone to JUSTIFY it

Relax, buddy.
Triangle ACD is congruent to triangle BDC, right?
The rest follows...