CDS 2026 (I) Elementary Mathematics, question 67

Topic: Geometry

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Q67

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=xAP=x, PC=yPC=y and BP=zBP=z, then what is (AC+BD)(AC+BD) equal to ?

  1. (a) (xy+yz+zx)z\frac{(xy+yz+zx)}{z}
  2. (b) (xy+yz+zx+z2)z\frac{(xy+yz+zx+z^{2})}{z}Correct answer
  3. (c) (xy+yz+zx+y2)y\frac{(xy+yz+zx+y^{2})}{y}
  4. (d) (xy+yz+zx+x2)x\frac{(xy+yz+zx+x^{2})}{x}

Solution.

Answer: (b) (xy+yz+zx+z2)z\frac{(xy+yz+zx+z^{2})}{z}

  1. Because ∠BAC\angle BAC and ∠BDC\angle BDC are right angles, A and D lie on the circle with diameter BC.
  2. The intersecting chords AC and BD therefore satisfy AP×PC=BP×PDAP \times PC = BP \times PD.
  3. Thus PD=xy/zPD = xy/z.
  4. With P inside both segments, AC=x+yAC = x + y and BD=z+xy/zBD = z + xy/z.
  5. Adding gives (xy+yz+zx+z2)/z\left( xy + yz + zx + z^{2} \right)/z.