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1. Titik O terlatak di tengah-tengah garis AB.
2. Titik P terletak di tengah tengah garis BD.
dan, AD =BD (Rusuk Tegak)
OD² = AD² - AO²
OD² = 18² - (½ 16)²
OD² = 324 - 64
OD² = 260
OD = √(4 .65)
OD = 2√65 cm
L∆ = L∆
½ AB . OD = ½ BD . AP
AB . OD = BD . AP
16 . 2√65 = 18 . AP
AP = (16/9)√65 cm