✔ 最佳答案
7.
答案是: (D) 4
∠CAQ = ∠DAQ(已知)
∠ACQ = ∠ADQ = 90° (已知)
AQ = AQ (公共邊)
所以 ΔACQ ≡ ΔADQ(AAS)
DQ = CQ = 1
2∠QAD + 2∠PAD = 180° (直線同側內角)
∠QAD + ∠PAD = 90°
但 ∠QAD + ∠AQD = 180° - ∠ADQ = 90°
故此 ∠PAD = ∠AQD
而且 ∠ADP = ∠QDA = 90° (已知)
所以 ΔPAD ~ ΔAQD (AAA)
DP/AD = AD/DQ (對應邊,ΔPAD ~ ΔAQD)
DP/2 = 2/1
DP = 4
∠PAB = ∠PAD (已知)
∠PBA = ∠PDA = 90° (已知)
PA = PA (公共邊)
ΔPBA ≡ ΔPDA (AAS)
BP = DP = 4 (對應邊,ΔPBA ≡ ΔPDA)
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2.
由於 PM 是 BC的垂直平分線:
∠BMP = 90° 及 BM= (1/2)BC = 3
在直角ΔPBM中:
BP² = PM² + BM² (勾股定理)
4² = PM² +3²
PM² = 7
PM = √7
作 AB 的垂直線 PN,與 AB相交於 N。
由於 P 點在 ∠ABC的角平分線上,PM = PN
BP = BP (公共邊)
∠BMP = ∠BNP = 90° (已知)
ΔBMP ≡ ΔBNP (RHS)
PN = PM = √7 (對應邊,ΔBMP ≡ ΔBNP)
BN = BM = 3 (對應邊,ΔBMP ≡ ΔBNP)
在直角ΔAPN中:
AN = AB - BN = (3 + √5) - 3 = √5
AP² = AN² +PN² (勾股定理)
AP² = (√5)² + (√7)²
AP² = 12
AP = 2√3