2026 · 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 = x, PC = y and BP = z, then what is (AC + BD) equal to?
Elementary MathematicsGeometry: Lines, Angles and Triangles
- A(xy + yz)/z
- B(xy + yz + zx + z²)/z
- C(xy + yz + zx + y²)/y
- D(xy + yz + zx + x²)/x
Explanation
Place A=(0,0), B=(b,0), C=(0,x+y), P=(0,x); from BP=z, b²+x²=z². The line through B and P meets the perpendicular-from-D condition DB⊥DC at D, giving BD=z+xy/z (via the parameter s=xy/z² along BP). Hence AC+BD=(x+y)+(z²+xy)/z=(xy+yz+zx+z²)/z, option B.