http://img40.imageshack.us/i/circle2.jpg/
my answer to part (i) is angle ADB is 90 degrees( angle in semicircle)
for part (ii) and (iii), i am lost, help is appreciated, thanks!
Originally posted by Divanhot:http://img40.imageshack.us/i/circle2.jpg/
my answer to part (i) is angle ADB is 90 degrees( angle in semicircle)
for part (ii) and (iii), i am lost, help is appreciated, thanks!
Since ADB = 90 and AB=AC. Therefore CD=DB ( perpendicular bisector)
Therefore DB/BC= 1/2
EB/GB = 1/2
Angle DBG = Angle CBE ( common angle)
Therefore Triangle DBG is similar to Triangle CBE. (ASS)
Therefore Angle CEB = Angle DGB ( corresponding angles of similar triangles)
Therefore CE and DG are parallel. ( Corresponding angles CEB and DGB)
Part iii) Using part ii) prove that triangle AFE is similar to triangle ADG.
Ratio of these triangles is 1:2.
Thus AF/AD = 1/2
AF = (1/2) AD shown.
Hi,
Approach to part ii:
1. Angle ADB = 90 degrees (from part i).
2. AB = AC (given).
3. Triangle ABC is isoceles (by 2).
4. BD = DC (by 1 & 3).
5. EG = GB (given).
6. DG // CE (by mid-point theorem).
Consider intercept theorem for part iii. Thanks!
Cheers,
Wen Shih
