Two triangles ABC right-angled at A and DBC right-angled at D are drawn such that AC and DB intersect at P. If AP = x, PC = y and BP = z, then what is (AC + BD) equal to ?

  1. A. \frac{(xy + yz + zx)}{z}
  2. B. \frac{(xy + yz + zx + z^2)}{z} ✓
  3. C. \frac{(xy + yz + zx + y^2)}{y}
  4. D. \frac{(xy + yz + zx + x^2)}{x}

Correct Answer: B. \frac{(xy + yz + zx + z^2)}{z}

Explanation

Since \angle A = \angle D = 90^\circ and \angle APB = \angle DPC (vertically opposite), \Delta APB \sim \Delta DPC. Therefore, \frac{AP}{DP} = \frac{BP}{CP} \implies \frac{x}{DP} = \frac{z}{y} \implies DP = \frac{xy}{z}. Then AC = x + y and BD = z + \frac{xy}{z}. Their sum is x + y + z + \frac{xy}{z} = \frac{zx + zy + z^2 + xy}{z}.

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