Discovering Triangle Symmetry: Tips for Solving Angle and Side Length Problems

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Sorry I am not sure if I should put this thread here.. looks like I am going to be told off by the boss :frown:

Anyway, here is my question in a triangle ABC, AB = AC and D is a point on AC such that AD = DB = BC. Find the size of the angle BAD? Find the angles of triangle ABC?

Just need a few hints to get this problem started please?
 
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You triangle, ABC, involves three related/interdepandent similar triangles.

Draw out the triangle(s). Try doing it to scale (i.e. try making the triangles look similar).
Label the angles. It should fall out pretty easily.
 
I have ADB being isoceles where AD=DB but from there how do you mesure angle A? I know that A and B will be the same in that triangle but how do u get a value?
 
There are three isosceles triangles. ABC and ADB are two of them. You have to find the third one before you can solve for the angles.
One property of isosceles triangles is that, not only are two sides of equal length, but the angles oppposite them are also equal.
Label all the (equal) angles and it will fall out.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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