Question

588 in the diagram below of isosceles triangle ahe with the vertex angle at h, \\(\overline{cb}\perp\overline{ae}\\) and \\(\overline{fd}\perp\overline{ae}\\).
which statement is always true?

1. \\(\dfrac{ah}{ac} = \dfrac{eh}{ef}\\)
2. \\(\dfrac{ac}{ef} = \dfrac{ab}{ed}\\)
3. \\(\dfrac{ab}{ed} = \dfrac{cb}{fe}\\)
4. \\(\dfrac{ad}{ab} = \dfrac{be}{de}\\)

Explanation:

Step1: Analyze the triangle properties

Since \( \triangle AHE \) is isosceles with \( AH = EH \) (vertex angle at \( H \)), and \( CB\perp AE \), \( FD\perp AE \), so \( \angle ABC=\angle EDF = 90^\circ \), and \( \angle A=\angle E \) (base angles of isosceles triangle). Thus, \( \triangle ABC\sim\triangle EDF \) (by AA similarity criterion).

Step2: Use similarity of triangles

For similar triangles \( \triangle ABC \) and \( \triangle EDF \), the ratios of corresponding sides are equal. That is, \( \frac{AC}{EF}=\frac{AB}{ED} \), because \( AC \) corresponds to \( EF \) and \( AB \) corresponds to \( ED \) in the similar triangles.

Step3: Check other options

• Option 1: \( \frac{AH}{AC}=\frac{EH}{EF} \), but \( AH = EH \), so this would imply \( AC = EF \), which is not necessarily true as \( C \) and \( F \) are arbitrary points on \( AH \) and \( EH \) respectively.
• Option 3: \( \frac{AB}{ED}=\frac{CB}{FE} \), but from similar triangles \( \frac{AB}{ED}=\frac{CB}{FD} \), not \( FE \), so this is false.
• Option 4: \( \frac{AD}{AB}=\frac{BE}{DE} \), there is no such corresponding side relationship from the given diagram and triangle properties.

Answer:

1. \( \boldsymbol{\frac{AC}{EF}=\frac{AB}{ED}} \)