Parallelograms and angles. PLEASE HELP

[mike2313]

Parallelogram ABCD has diagonal AC equal in length to side AB. CD is produced to E so that D is between E and C. If angle BAC = 30 degrees, find the size of angle ADE

 

Please Help

 

Thanks in advance

[FractalFiend]

Smells like homework! Sorry i won't hurt your intellectual growth by doing your homework for you. Goodluck!

[Rajdilawar S]

Your answer is 150 degree.

Consider Triangel ABC

Since AB=AC, therefore it is a issoseles triangel.Therefore angle ACB=ABC=60 degree.And since angle BAC = 30 degree (as angle BAC is given 30 degree.)

Also in Parallelograms ABCD AB = DC, and given that AB=AC.

Therefore angle ADC=30 degree.

So angle ADE = 180-30=150 degree.

Draw the figure.

[5614]

Therefore angle ACB=ABC=60 degree

WHAT?! No it doesn't. You can just look at it to see that ABC must be greater than 90.

 

Here is the diagram I got:

 

 

Proof 1:

 

Since CAB=30 we can say DAB=60

 

As DAB=DCB=60 and all the angles added together sum to 360 we can say that:

ADC + ABC = 360 - DAB - DCB

ADC + ABC = 360-60-60

ADC + ABC = 240

And as ADC=ABC they both equal 240/2 = 120

 

ADE = 180 - ADC

ADE = 180 - 120

ADE = 60

 

Proof 2:

 

Now are you familiar with alternate angles?

 

So CAB=30 therefore DAB=60

 

Now using alternate angles if DAB=60 then ADE=60

 

A diagram showing a similar use of alternate angles is here:

http://www.gomath.com/geometry/parallelogram.jpg

it is the alpha angle we are interested in on the right hand side.

[hgupta]

Since CAB=30 we can say DAB=60

How??

 

it should be 105^o.

 

taking tri(ABC), given ang(BAC) = 30 and AB = BC, so ang(ACB) = ang(ABC).

 

sum of all angles of a triangle = 180^o

 

therefore, ang(ABC+ACB+BAC) = 180^o and BAC = 30^o

=> 2*ang(ACB) = 150^o since ACB = ABC

=> ACB = 75^o

 

=> ang(CAD) = ang(ACB) = 75^o, property of ||ogram, alternate angles of a diagnol are equal

 

=> BAD = BAC + CAD = 30^o + 75^o = 105^o

 

also, BAD = ADE, property of || lines

 

therefore, ADE = 105^o

 

WAIT, could you please re-read and re-post your question!! if AC = AB then it is no longer a ||gram but a quadrilateral!!

[rajama]

Parallelogram ABCD has diagonal AC equal in length to side AB. CD is produced to E so that D is between E and C. If angle BAC = 30 degrees' date=' find the size of angle ADE

 

Please Help

 

Thanks in advance[/quote']

 

It may smell like homework but two thirds of the answers so far are a little disappointing... lets see if I can't make things worse...

 

AC = AB & BAC = 30' : ABC isosceles triangle pointing top left

ABC = (180' - 30')/2 = 75'

parallelogram : ABC = CDA = 75'

ADE = 180' - CDA = 105'

 

This appears to confirm the hgupta result - congratulations - and we can all go home...

 

(quietly: I think all the confusion was deliberate wasn't it, to discourage homework questions, hmm? )

[The Thing]

5614. It's the other diagonal. With your diagonal Triangle ABC is obtuse, which means that AB does not equal AC, contrary to what's given. Proof of that is quite simple: Isosceles triangles have two equal base angles, namely angles ACB and ABC here. If ABC is greater than 90, ACB is greater than 90 and there's the whole 180 degree in the triangle theorem.

 

So yeah, draw the shorter diagonal, you find Angle ABC is (180-30)/2=75. I think Rajdilawar just had an arithmetic error =]. Yeah then opposite angles of a parallelogram are same then supplementary angles. So it is 105 just to confirm.

 

Now I'm REALLY sure Rajdilawar just had a calculation error.

[5614]

5614. It's the other diagonal. With your diagonal Triangle ABC is obtuse, which means that AB does not equal AC, contrary to what's given.

Ah yeah, my diagram was wrong, hence my calculations were too. Sorry, thanks for pointing it out!