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已知 \newline△ABC\triangle ABC 中, \newlineAC=8AC=8,AB=41AB=\sqrt{41},BCBC 边上的高 \newlineAG=5AG=5,DD 为线段 \newlineACAC 上的动点, 在 \newlineBCBC 上截取 \newlineCE=ADCE=AD, 连接 \newlineAE,BDAE,BD, 则 \newlineAC=8AC=800 的最小值为\newlineAC=8AC=811 题图

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Q. 已知 \newline△ABC\triangle ABC 中, \newlineAC=8AC=8,AB=41AB=\sqrt{41},BCBC 边上的高 \newlineAG=5AG=5,DD 为线段 \newlineACAC 上的动点, 在 \newlineBCBC 上截取 \newlineCE=ADCE=AD, 连接 \newlineAE,BDAE,BD, 则 \newlineAC=8AC=800 的最小值为\newlineAC=8AC=811 题图
  1. Calculate Area of Triangle: Calculate the area of triangle ABC using the height AG and base BC. \newlineArea = 12×base×height=12×AC×AG=12×8×5=20\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times AG = \frac{1}{2} \times 8 \times 5 = 20.
  2. Use Heron's Formula: Use Heron's formula to find the semi-perimeter ss of triangle ABC.s=AC+AB+BC2=8+41+BC2s = \frac{AC + AB + BC}{2} = \frac{8 + \sqrt{41} + BC}{2}.We don't know BCBC yet, so we can't calculate ss at this moment.
  3. Express BC in Terms: Use the area we found to express BC in terms of the other sides.\newlineArea = s⋅(s−AC)⋅(s−AB)⋅(s−BC)\sqrt{s \cdot (s - AC) \cdot (s - AB) \cdot (s - BC)}.\newlinePlug in the known values and solve for BC.\newline20=s⋅(s−8)⋅(s−41)⋅(s−BC)20 = \sqrt{s \cdot (s - 8) \cdot (s - \sqrt{41}) \cdot (s - BC)}.\newlineWe have two unknowns here, ss and BCBC, so we can't solve this directly.
  4. Correct Mistake: Notice that we made a mistake in the previous step; we can't use Heron's formula without knowing all three sides.\newlineWe need to find BCBC using the area and the height AGAG.\newlineArea =12×BC×AG= \frac{1}{2} \times BC \times AG, so BC=2×AreaAG=2×205=8BC = \frac{2 \times \text{Area}}{AG} = \frac{2 \times 20}{5} = 8.

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