Circle( A Maths)

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!

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