The perpendicular AD on the base BC of a triangle ABC intersects BC at D so that DB=3\text{ CD}. Which one of the following is correct?

  1. A. 2(AB+AC)(AB-AC)=BC^{2} ✓
  2. B. 3(AB+AC)(AB-AC)=2BC^{2}
  3. C. 4(AB+AC)(AB-AC)=3BC^{2}
  4. D. 5(AB+AC)(AB-AC)=4BC^{2}

Correct Answer: A. 2(AB+AC)(AB-AC)=BC^{2}

Explanation

In right \triangle ABD and \triangle ACD, AD^2 = AB^2 - BD^2 = AC^2 - CD^2 \implies AB^2 - AC^2 = BD^2 - CD^2. Given BD = 3CD, the base BC = 4CD, so CD = \frac{BC}{4} and BD = \frac{3BC}{4}. Substituting gives AB^2 - AC^2 = \frac{9BC^2}{16} - \frac{BC^2}{16} = \frac{BC^2}{2}. Thus, 2(AB^2 - AC^2) = BC^2, or 2(AB+AC)(AB-AC) = BC^2.

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