✔ 最佳答案
AB // CD (mid-pt. theorem)
∠ AJB = ∠ CJD (vert. opp.∠s)
∠ JAB = ∠ JCD (alt.∠s, AB//CD)
∴ △CJD ~ △AJB(A.A.)
∴ AJ : JC = BJ : JD
22.5 : 9 = (JH + HB) : JD
5 : 2 = (JH + HB) : (DH - JH)
Since DH = HB = 14 (mid-pt. theorem)
∴
5 : 2 = (JH + 14) : (14 - JH)
2(JH + 14) = 5(14 - JH)
2JH + 28 = 70 - 5JH
7JH = 42
JH = 6