I need some help with this math prob.

[cali.us.101]

ABCD is an isosceles trapezoid with line AD congruent to line BC. Diagonal line AC bisects angle DAB. If angle CAB = 15, find:
a) angle B b) angle BCA c) angle ACD d) angle D

 

[tkhunny]

Let's see some effort. Is there a drawing? Is it JUST a trepezoid or could it me, say, a parallelogram, too?

If you can find Angle DAC, we can have more discussion.

 

[BigGlenntheHeavy]

\(\displaystyle Is \ this \ what \ you \ want?\)

[attachment=0:26a5f1e6]tri.JPG[/attachment:26a5f1e6]

 

[cali.us.101]

no there's not picture. everything i wrote is all it says.

 

[Deleted member 4993]

no there's not picture. everything i wrote is all it says.

Okay, so Glenn gave you a drawing. Is it correct according to your problem statement?

If it is correct - what can you work out regarding the given problem?

If it is not correct - fix it and tell us - what can you work out regarding the given problem?

 

[cali.us.101]

i think the picture is accurate.

 

[Deleted member 4993]

i think the picture is accurate.

Okay then work with it!

How much is angle CAD - knowing Ac bisects DAB?

How much is angle DCA - knowing AC||DC?

How much is angle BAD - knowing AC bisects DAB?

 

[cali.us.101]

i don't know. i've been working on this problem for months. i don't know what each is angle is.

 

[BigGlenntheHeavy]

\(\displaystyle cali, \ if \ you \ have \ a \ basic \ grasp \ of \ geometry, \ there \ shouldn't \ be \ any \ problem.\)

\(\displaystyle However, \ if \ you \ been \ pondering \ this \ problem \ for \ months, \ then \ a \ good \ review \ of \ the \ basics \ is\)

\(\displaystyle in \ order.\)

 

[cali.us.101]

how can i solve this problem if i don't knw the value os angle BCA?

 

[BigGlenntheHeavy]

\(\displaystyle Ok, \ I'll \ do \ this \ one \ for \ you.\)

\(\displaystyle Angle DAC \ = \ 15^0, \ half \ of \ bisected \ angle \ A.\)

\(\displaystyle Angle \ DCA \ = \ 15^0, \ alternate \ interior \ angles \ cut \ by \ a \ tranversal \ are \ =, \ DC \ is \ parallel \ to \ AB.\)

\(\displaystyle Hence, \ angle \ D \ = \ 150^0, \ Triangle \ = \ 180^0\)

\(\displaystyle Angle \ ABC \ = \ 30^0, \ Isosceles \ Trapezoid, \ AD \ is \ congruent \ to \ BC, \ given.\)

\(\displaystyle Therefore, \ angle \ ACB \ = \ 180^0-(30^0+15^0) \ = \ 135^0.\)

\(\displaystyle Note: \ This \ is \ one \ of \ a \ plethora \ of \ ways \ to \ solve \ the \ above \ problem, \ it \ isn't \ unique.\)