如图, BCBC 是以 OO 为圆心的圆的直径。HH 是 CFCF 的中点。
ABCABC 是一条直线,且 AGAG 是圆在点 FF 处的切线。直线 EFEF 与 OHOH 相交于点 XX,EE 是 COCO 的中点。

  1. (i)

    证明三角形 ABFABF 和 AFCAFC 相似。

  2. (ii)

    证明 AF2=AB2+AB×BCAF^2 = AB^2 + AB \times BC。

  3. (iii)

    证明 OX:XH=2:1OX : XH = 2 : 1。

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完整解法

13(i)

1. Identify a shared angle between the two triangles

In △ABF\triangle ABF and △AFC\triangle AFC, ∠FAB=∠FAC\angle FAB = \angle FAC (common angle).

2. Use the alternate segment theorem to find a second pair of equal angles

Since AGAG is a tangent to the circle at FF and FCFC is a chord, by the alternate segment theorem, ∠AFB=∠ACF\angle AFB = \angle ACF.

3. Conclude similarity using the AA criterion

Since two pairs of corresponding angles are equal, △ABF\triangle ABF is similar to △AFC\triangle AFC (AA similarity).

答案

Triangles ABFABF and AFCAFC are similar (AA similarity).

13(ii)

1. Use the similarity of triangles from part (i) to set up a ratio of corresponding sides.

Since △ABF\triangle ABF is similar to △AFC\triangle AFC, the ratio of their corresponding sides is equal:
ABAF=AFAC\frac{AB}{AF} = \frac{AF}{AC}

2. Express $AF^2$ in terms of $AB$ and $AC$, then substitute $AC$ using segment addition.

Cross-multiplying gives:
AF2=AB×ACAF^2 = AB \times AC
Since A,B,A, B, and CC lie on a straight line in that order, AC=AB+BCAC = AB + BC. Substituting this into the equation:
AF2=AB(AB+BC)AF^2 = AB(AB + BC)
AF2=AB2+AB×BCAF^2 = AB^2 + AB \times BC

答案

AF2=AB2+AB×BCAF^2 = AB^2 + AB \times BC (Shown)

13(iii)

1. Apply the Midpoint Theorem to triangle COF.

In △COF\triangle COF, EE is the midpoint of COCO and HH is the midpoint of CFCF.
By the Midpoint Theorem, EH∥OFEH \parallel OF and EH=12OFEH = \frac{1}{2}OF.

2. Identify equal angles to prove similarity between triangles OXF and HXE.

Consider △OXF\triangle OXF and △HXE\triangle HXE:
∠OXF=∠HXE\angle OXF = \angle HXE (vertically opposite angles)
∠XOF=∠XHE\angle XOF = \angle XHE (alternate angles, EH∥OFEH \parallel OF)
∠XFO=∠XEH\angle XFO = \angle XEH (alternate angles, EH∥OFEH \parallel OF)

3. State the similarity of the triangles.

Therefore, △OXF\triangle OXF is similar to △HXE\triangle HXE.

4. Use the ratio of corresponding sides to find the ratio of OX to XH.

Since the triangles are similar, the ratio of their corresponding sides is equal:
OXXH=OFEH\frac{OX}{XH} = \frac{OF}{EH}
Since EH=12OFEH = \frac{1}{2}OF, we have OFEH=21\frac{OF}{EH} = \frac{2}{1}.
Thus, OXXH=21\frac{OX}{XH} = \frac{2}{1}, which means OX:XH=2:1OX : XH = 2 : 1.

答案

Shown that OX:XH=2:1OX : XH = 2 : 1.

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