Analysis: Construct the golden section point C of the line segment AB, so that AB:AC=AC:CB, that is, AC^2=AB*CB,
Method: Conclude BD⊥AB in B and BD=1/2AB, connect AD, take point E on AD to make ED=BD, take point C on AB to make AC=AE, then point C is the golden section point of line segment AB
Proof: Connect BE, assuming AB=1, then BD=1/2, AD=√(AB^2+BD^2)=√[1^2+(1/2)^2]=√5/2< /p>
AC=AE=AD-DE=√5/2-1/2, CB=1-AC=1-(√5/2-1/2)=3/2-√5/2
∴AC^2=(√5/2-1/2)^2=3/2-√5/2, AB*CB=1*(3/2-√5/2) =3/2-√5/2
That is, AC^2=AB*CB, ∴ point C is the golden section point of the desired line segment AB.
(Because AC^2=AB*CB, that is, AC is the middle term in the ratio of AB and CB, so point C is also called the dividing line segment AB into the middle and outer ratio. If AB=1, then AC=√5/ 2-1/2≈0.618)