Simple geormetry/congruence help

ABCD is a quadrilateral with AB horizontal and CD vertical. Angle BAD = angle BCD = 40 degrees and AB=BC=5. Show that angle CDB = angle ADB = 25 degrees.

Here is a diagram I made which I hope is correct:

http://img443.imageshack.us/img443/4404/geom1.jpg

By observing alternate angles, I can see that angle ADC = 50 degrees but how do I show that angle CDB = angle ADB? I thought about proving congruence but is this possible here?

Let's analize triangles ADB and BDC:
- angle DAB = angle BCD
- BD is a common lature
- AB = BC = 5
Rezults from the case of congruence side-side-angle triangle ADB= triangle BDC so angleADB=angleBDC

AB is horizontal and CD is vertical, so if we extend the side AB we'll have a right angle where AB intersect CD (point F). In triangle AFD, F=90 degrees, A=40 degrees, so angleADF = angleADC = 180-90-40 = 50 degrees.

AngleADB is half of angleADC, therefore is 25 degrees.

Quote:

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Originally Posted by draganicimw

Let's analize triangles ADB and BDC:
- angle DAB = angle BCD
- BD is a common lature
- AB = BC = 5
Rezults from the case of congruence side-side-angle triangle ADB= triangle BDC so angleADB=angleBDC

AB is horizontal and CD is vertical, so if we extend the side AB we'll have a right angle where AB intersect CD (point F). In triangle AFD, F=90 degrees, A=40 degrees, so angleADF = angleADC = 180-90-40 = 50 degrees.

AngleADB is half of angleADC, therefore is 25 degrees.

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But I thought SSA by itself isn't enough to prove congruence?

I agree, side, side, non-incuded angle does not imply congruence (though in fact they are).
You can get there by extending AB to meet CD at E and CB to meet AD at F.
Deduce that the angle at F is a rightangle and then that triangles AFB and CEB are congruent and finally that triangles DFB and DEB (rtangle hypotenuse side) are congruent.
As an alternative you could use trig, the sine rule in triangles ADB and CBD.