In figure 3.95, chord EF ∥ chord GH. Prove that, chord EG ≅ chord FH. Fill in the blanks and write the proof. Proof: Draw seg GF. ∠ EFG = ∠ FGH ………. □ (I) ∠ EFG = □ ……. inscribed angle theorem (II) ∠ FGH = □ ……. inscribed angle theorem (III) therefore m(arc EG) = □ from (I), (II), (III). chord EG ≅ chord FH ………. □

Answer: Alternate angles ∠ EFG = ∠ FGH give m(arc EG) = m(arc FH), so chord EG ≅ chord FH. Draw seg GF. ∠ EFG = ∠ FGH …… alternate angles (EF ∥ GH, GF is a transversal) (I) ∠ EFG = 1/2 m(arc EG) …… inscribed angle theorem (II) ∠ FGH = 1/2 m(arc FH) …… inscribed angle theorem (III) therefore m(arc EG) = m(arc FH) from (I), (II), (III). chord EG ≅ chord FH …… chords corresponding to congruent arcs are congruent.

In figure 3.95, chord EF ∥ chord GH. Prove that, chord EG ≅ chord FH. Fill in the blanks and write the proof.

Proof: Draw seg GF.

\(\angle EFG = \angle FGH\) ………. □ (I)

\(\angle EFG\) = □ ……. inscribed angle theorem (II)

\(\angle FGH\) = □ ……. inscribed angle theorem (III)

\(\therefore\) m(arc EG) = □ from (I), (II), (III).

chord EG ≅ chord FH ………. □

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Answer: Alternate angles \(\angle EFG = \angle FGH\) give m(arc EG) = m(arc FH), so chord EG ≅ chord FH.

Draw seg GF.

\(\angle EFG = \angle FGH\) …… alternate angles (EF ∥ GH, GF is a transversal) (I)

\(\angle EFG = \frac{1}{2}\) m(arc EG) …… inscribed angle theorem (II)

\(\angle FGH = \frac{1}{2}\) m(arc FH) …… inscribed angle theorem (III)

\(\therefore\) m(arc EG) = m(arc FH) from (I), (II), (III).

chord EG ≅ chord FH …… chords corresponding to congruent arcs are congruent.

Doubtly AI · 24 Oct 2026 AI-verified · Oct 2026 Report

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In figure 3.95, chord EF ∥ chord GH. Prove that, chord EG ≅ chord FH. Fill in the blanks and write the proof.

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