Geometry problem,very intersting, please check!!

[zafer]

1. The problem statement, all variables and given/known data
In the figure,AH=HC and AB=DC,measure of angle A is 54.Whats the measure of the angle C?
http://img685.imageshack.us/img685/9144/gty.png

2. Relevant equations
Isosceles triangle theorem, altitudes ,medians ,angle bisectors...


3. The attempt at a solution
I have tried to extend a line segment A to D which is equal to the line segment DC.
Now we have an Isosceles triangle DAC.Since we now that in an isosceles triangle the altitude which is drawn from the vertex of the triangle is also median and angle bisector.So the angle D or the vertex is bisected in to two congruent angles.
And if to sides are equal DA with DC and also the angles opposite them should be equal.
We can name the angles which are congruent as ones that are congruent with the letter a and the others with letter b.The sum of these will be a+b=90.
And then I don't know what to do???

[Mark44]

1. The problem statement, all variables and given/known data
In the figure,BH=HC and AB=DC,measure of angle B is 54.Whats the measure of the angle C?
What is all the given information? In your figure, you show DH = HC, not BH = HC, and your drawing shows angle DHC as a right angle, but that is not stated in the given information. The picture seems to show that AB and DC are congruent, but that is not given information.

http://img685.imageshack.us/img685/9144/gty.png

2. Relevant equations
Isosceles triangle theorem, altitudes ,medians ,angle bisectors...


3. The attempt at a solution
I have tried to extend a line segment A to D which is equal to the line segment DC.
Now we have an Isosceles triangle DAC.Since we now that in an isosceles triangle the altitude which is drawn from the vertex of the triangle is also median and angle bisector.So the angle D or the vertex is bisected in to two congruent angles.
And if to sides are equal DA with DC and also the angles opposite them should be equal.
We can name the angles which are congruent as ones that are congruent with the letter a and the others with letter b.The sum of these will be a+b=90.
And then I don't know what to do???

[zafer]

No no, AB is congruent to DC and AH is congruent to HC.

Sorry, I gave the wrong information.
If i extend a line segment from A to D will it be congruent to DC

[Mark44]

What about angle DHC? Is that a right angle? Your picture shows it as one, but that's not in the given information, which you have changed. It's very difficult to provide help if the information changes or is incomplete.

For the second time, please state all of the information that is given.

[jack action]

What you need to do is to draw another vertical line from point B to the base line AC (let's call this new point E).

You will have 2 new rectangle triangles ABE and BEC.

Looking at BEC:

You can find BE length easily with angle A and length AB

You can find EC length easily with AC = 2*HC, angle A and length AB

Knowing that BEC and DHC share the same angle C, you can find a relationship between them.
This final relationship will be (there are a lot of manipulations to get there):

sin 54 / sin x + cos 54 / cos x = 2

(To get x, I solved it by trial and error and plotting the function, I don't know if there is an easier way)

This equation gives you 2 good answers for x. The first one is x = 54°, which means that BE = DH, such that there are the exact same line (BE), which is not your case, obviously. The other answer has DH < BE, which is your case (actually DH = 0.827*BE).

[willem2]

sin 54 / sin x + cos 54 / cos x = 2

(To get x, I solved it by trial and error and plotting the function, I don't know if there is an easier way)



multiply with sin x * cos x .
then the left side becomes sin(x+54) and the right side sin(2x)

[jack action]

multiply with sin x * cos x .
then the left side becomes sin(x+54) and the right side sin(2x)

And then with further transformations, you get the other solution:

sin(-x+126) = sin(2x)

Trigonometry transformations are far, but it's all coming back now!