CSSA 2011 Question 6 d) iii (1 Viewer)

[Dubble-u25]

I can't find the last proof I need to prove that the triangles are congruent. Judging by the marks it should be quite simple but I can't see it.

In trianlge CAB and trianlge GBA:
Angle ACB = angle AGB (angle on the circumference subtended from the same arc AB are equal)
AB = AB (common side)
???

 

[someth1ng]

I can't see the proof either...but I can't draw all over it the circle...

 

[michaeljennings]

I can't find the last proof I need to prove that the triangles are congruent. Judging by the marks it should be quite simple but I can't see it.

In trianlge CAB and trianlge GBA:
Angle ACB = angle AGB (angle on the circumference subtended from the same arc AB are equal)
AB = AB (common side)
???
View attachment 23227

^CAB=180-2(alpha) - angles on a straight line supplementary
^GBA=180-2(alpha) - angle sum of triangle DBG

 

[someth1ng]

^CAB=180-2(alpha) - angles on a straight line supplementary
^GBA=180-2(alpha) - angle sum of triangle DBG

Yeah...didn't see that, well, I saw half of it. lol

 

[xxxzxc]

third angle proof:
cos aed is isosceles the apex (angle aeg) is 180-2a
angle cab is 180-2a (supplementary angles)
angle abg = angle aeg (angles subtended by same arc are equal)
hence, triangle ACB is congruent to AGB (AAS)
the trick is finding the other angle cos they dont give you hints for any other side

 

[Dubble-u25]

Ohhhh.... i see now thanks