✔ 最佳答案
(a)
∠OCQ = ∠ACD(common ∠)
∠OQC = 90° (int. ∠ ofsquare)
∠ADC = 90° (radius ⊥tangent)
Hence, ∠OQC= ∠ADC
ΔCOQ~ ΔCAD (AAA)
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(b)
CP = CD = h cm (radii of same circle)
ΔCOQ ~ ΔCAD (proven)
CO/CA = OQ/AD (corr. sides of similar Δs)
CO/CA = k/h
(CP - OP)/(AP + CP) = k/h
(h - k)/(AP + h) = k/h
k(AP + h) = h(h - k)
kAP + hk = h² - hk
kAP = h² - 2hk
kAP = h(h - 2k)
AP= h(h - 2k)/k
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(c)
Join QS and PQ. (construction)
∠QCS = ∠PCQ(common ∠)
∠CQS = ∠CPQ(∠s in alt. segment)
Hence, ΔCQS ~ ΔCPQ (AAA)
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(d)
ΔCQS ~ ΔCPQ (proven)
CQ/CP = CS/CQ (corr. sides of similar Δs)
k/h = (h - 2k)/k
Hence, k² = (h - 2k)h
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(e)
From (b) :
AP = h(h - 2k)/k ...... [1]
From (d) :
k² = (h - 2k)h ...... [2]
[1]/[2] :
AP/k² = 1/ k
AP = k