Can you help me to solve this problem ?

The answer must be just by geometric prove .Not by trigonometric !!!

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- Nov 5th 2010, 05:33 AMyehoramHow much is alpha ?
Can you help me to solve this problem ?

The answer must be just by geometric prove .Not by trigonometric !!! - Nov 5th 2010, 06:14 AMUnknown008
From the figure, we know that $\displaystyle \angle ABC = \angle ACB = 50^o$

So, $\displaystyle \angle DCB = 30^o$

From triangle ACD, we know that $\displaystyle \angle ADC = 150^o$

So, we get some equations now.

Let's give some names to the other angles:

$\displaystyle \angle ABD = a$

$\displaystyle \angle DBC = b$

$\displaystyle \angle BDC = c$

From triangle ABD, we know that: $\displaystyle 70^o + \alpha + a = 180^o$

From triangle DBC, we know that: $\displaystyle b + c + 30^o = 180^o$

And lastly, we know that $\displaystyle \alpha + 150^o + c = 360^o$

and $\displaystyle b + c = 50^o$

There is enough information to find alpha (Smile) - Nov 6th 2010, 04:57 AMOpalg
Unfortunately, that is not enough information to find alpha. There are four equations for four unknowns, but they are not independent. (The fourth equation, in its corrected form above, is just what you get by adding the first two equations and subtracting the third.)

By trigonometry I get the answer to be that $\displaystyle \alpha = 70^\circ$, but I cannot see any way of proving that by synthetic geometry. This seems to be a variant of the notorious 80-80-20 triangle problem, but maybe this one is even harder. - Nov 6th 2010, 05:33 AMUnknown008
You're right... that's what happens when I had a diagram with alpha, beta, gamma, delta, etc and making a post with other letters... (Doh)

- Nov 6th 2010, 09:28 AMWilmer