Can you help me finding

the angle A (see the attachment please.)

Attachment 20471

I don't know how to start ...

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- Jan 17th 2011, 03:43 AMrazemsoft21Find the angle of a triangle ...
Can you help me finding

the angle A (see the attachment please.)

Attachment 20471

I don't know how to start ... - Jan 17th 2011, 07:29 AMWilmer
Well, here's a start!

A = angle A : so angle C = 100 - A which leaves angle BDC = 40 + A

a = BC, b = AC, c = AB, d = BD : so AD = a + d

Insert point E on AD such that DE = d, which leaves AE = a :

so we now have isosceles triangle BDE, equal sides = d, equal angles = (40 + A) / 2.

Insert point F on AB, such that EF is parallel to BC :

so we have triangle AFE, similar to triangle ABC

I think that should lead you to the solution: give it a shot... - Jan 17th 2011, 12:57 PMrazemsoft21
All understood

but similar triangles treats the sides of

the triangle, how can I treat the angles ? - Jan 17th 2011, 01:41 PMpickslides
Well it seems ABC is isosceles, are you told that?

If so Angles ABC = 80 and BCA = 80

Angle BAC = 180 - (ABC - BCA) - Jan 17th 2011, 02:09 PMrazemsoft21
- Jan 17th 2011, 02:35 PMemakarov
Practicing my diagram skills:

http://lh6.ggpht.com/_SNa3egOo9rk/TT...ty-degrees.png

I agree that it is not obvious that AB = AC. Because of this, I think a crucial element of the solution is still missing. In particular, it is not clear how Wilmer's solution uses the fact that AE = a. - Jan 17th 2011, 03:51 PMbjhopper
I think the condition AD=DC+BCis only true for an isosceles triangle 20-80-80 with a specific base lenght

bjh - Jan 17th 2011, 04:13 PMrazemsoft21
- Jan 17th 2011, 04:13 PMemakarov
When two triangles are similar, every length is multiplied by a constant factor. So the relationship AD = BC + BD is preserved.

- Jan 18th 2011, 03:53 AMbjhopper
Yes that was a faulty assumption.I calculated the value of AD,BD,DC and BC for a triangle where AB = 10.BC =3.947 DC =2.578, AD=7.422 BD=3.949

7.422=3.947 +3.473

These values were calculated using trig. BC using cos 80/10 =1/2 of BC

using the angle bisector theorem BC/DC= 10/AD and from this 3.473 /DC = 10/10-DC

BD is found using the sin rule AC/sin60= BD/sin 80

Is there a geometry proof? I don't know

bjh - Jan 18th 2011, 05:04 AMArchie Meade

(1) Locate Q such that triangles BQD and BCD are congruent.

(2) Rotate triangle BCD about D until B is at P and C is at Q.

P, D and C will be collinear only if $\displaystyle \alpha=180^o-2\alpha\Rightarrow\ \alpha=60^o$

$\displaystyle |\angle\ DQA|=\alpha+40^o\Rightarrow\ |\angle\ QAP|=180^o-\left[180^o-2\alpha+\alpha+40^o\right]=\alpha-40^o$

Therefore A, Q and B are collinear and triangle APQ is isosceles only if $\displaystyle \alpha=60^o$

$\displaystyle \Rightarrow\ |\angle\ QPA|=140^o\Rightarrow\ |\angle\ QAP|=20^o$ - Jan 18th 2011, 09:11 AMrazemsoft21
- Jan 18th 2011, 10:05 AMemakarov
Something I don't understand...

At this point we don't know if it is possible for the rotated image of C to lie on AB*and*the image of B to lie on AC. Since above you defined Q but not P, I assume that we rotate BCD until C coincides with Q. The image of B, i.e., P, does not have to be on AC.

Quote:

P, D and C will be collinear only if $\displaystyle \alpha=180^o-2\alpha\Rightarrow\ \alpha=60^o$

Quote:

$\displaystyle |\angle\ DQA|=\alpha+40^o\Rightarrow\ |\angle\ QAP|=180^o-\left[180^o-2\alpha+\alpha+40^o\right]=\alpha-40^o$

Quote:

Therefore A, Q and B are collinear and triangle APQ is isosceles only if $\displaystyle \alpha=60^o$

Quote:

Originally Posted by**bjhopper**

Quote:

cos 80/10 =1/2 of BC

Quote:

from this 3.473 /DC = 10/10-DC

I also came up with a trigonometry solution. Let $\displaystyle \angle BAC=\alpha$. By the law of sines for BAD, $\displaystyle \displaystyle\frac{BD}{\sin\alpha}=\frac{AD}{\sin 40}=\frac{BD+BC}{\sin 40}$. Also, for BDC, $\displaystyle \displaystyle\frac{BD}{\sin(100-\alpha)}=\frac{BC}{\sin(40+\alpha)}$. Now, we may give BD any numeric value; then BC and $\displaystyle \alpha$ will be determined uniquely because the whole picture can be scaled arbitrarily preserving $\displaystyle \alpha$. So, let BD = 1 and denote BC by a. From the first equation, $\displaystyle a=(\sin40)/(\sin\alpha)-1$; from the second one, $\displaystyle a=(\sin(40+\alpha))/(\sin(100-\alpha))$. Therefore,

$\displaystyle \displaystyle f(\alpha):=\frac{\sin40}{\sin\alpha}-\frac{\sin(40+\alpha)}{\sin(100-\alpha)}=1$

It is easy to see that on (0, 100), the function $\displaystyle f(\alpha)$ is strictly decreasing: the denominator of the first fraction and the nominator of the second one increase while the denominator of the second fraction decreases. Therefore, there is at most one solution. With a bit of trig, it is possible to show that f(20) = 1. - Jan 18th 2011, 10:29 AMArchie Meade
- Jan 18th 2011, 10:55 AMemakarov
Sorry, I spent a lot of time on this, and I am still confused. Could you write more precisely how and in which order you define various objects and what you know about them at that time, not what you would like to prove eventually?

Quote:

At this point we don't know if it is possible for the rotated image of C to lie on AB

This we do know as |QD|=|DC| since triangles DBC and DQB have x, y and 40 degrees as shown.

Then we check the condition required for P to be on AC.

**or**that the image of B lies on AC. I thought you chose the first thing, making Q the image of C by definition. That's fine. But the rest of the proof uses the fact that the image of B (I presume, it is P by definition) is on AC, and I don't see it proved.