1. Geometry proof?

I have this Geometry problem here which is a proof question.
I am really unsure of where to start, but I have been told by the teacher that using congruent triangles might help, but does that mean I will need to draw in a line onto the diagram? I'm not quite sure where to start.
Thanks.

2. Let $\displaystyle \widehat{ABC}=\alpha, \widehat{BAC}=2\alpha, \widehat{ACD}=\widehat{DCB}=\beta$

In the triangle BDC we have $\displaystyle CD=\frac{BC\sin\alpha}{\sin(\alpha+\beta)}$

In the triangle ADC we have $\displaystyle CD=\frac{AD\sin 2\alpha}{\sin\beta}$

Then $\displaystyle \frac{BC\sin\alpha}{\sin(\alpha+\beta)}=\frac{2AD\ sin\alpha\cos\alpha}{\sin\beta}\Rightarrow BC\sin\beta=2AD\sin(\alpha+\beta)\cos\alpha\Righta rrow$

$\displaystyle \Rightarrow BC\sin\beta=AD[\sin(2\alpha+\beta)+\sin\beta]\Rightarrow BC=\frac{AD\sin(2\alpha+\beta)}{\sin\beta}+AD=$

$\displaystyle =\frac{AD\sin(\alpha+\beta)}{\sin\beta}+AD=AC+AD$

3. Originally Posted by f1f2f3
I have this Geometry problem here which is a proof question.
I am really unsure of where to start, but I have been told by the teacher that using congruent triangles might help, but does that mean I will need to draw in a line onto the diagram? I'm not quite sure where to start.
Thanks.
I wonder if this is the construction that the teacher had in mind. Let E be a point on BC such that angle EDC = angle ADC. Then see if you can prove these facts:
1. Triangles DEC, DAC are congruent.
2. Angle DEC = angle DAC = 2× angle DBE.
3. Triangle BDE is isosceles.
4. BE = DE = DA.
5. BC = BE + EC = AD + AC.

4. Sorry, I'm not quite sure how I'm supposed to prove the first step. How are triangles DEC and DAC congruent?

5. Originally Posted by f1f2f3
Sorry, I'm not quite sure how I'm supposed to prove the first step. How are triangles DEC and DAC congruent?
• The side CD is common to both triangles.
• Angle EDC = angle ADC (by construction).
• Angle ECD = angle ACD (because CD bisects angle ACB).
Therefore the triangles are congruent (one is a mirror image of the other).

6. Also, how do I know that triangle BDE is isosceles?

7. Originally Posted by f1f2f3
Also, how do I know that triangle BDE is isosceles?
The external angle DEC is the sum of the internal angles DBE + BDE. But we know that angle DEC = angle BAC, which we are told is equal to twice angle DBE. So angle DBE is half of angle DEC, and therefore so is angle BDE. Thus the angles DBE and BDE are equal.