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?

Since both triangles share hypotenuse BC and are right-angled at A and D respectively, both A and D lie on circle with BC as diameter. By intersecting chords theorem at P: AP × PC = DP × PB, so x·y = DP·z, giving DP = xy/z. Thus BD = BP + PD = z + xy/z = (z² + xy)/z. And AC = AP + PC = x + y. So AC + BD = (x + y) + (z² + xy)/z = (xz + yz + z² + xy)/z. Hmm, this gives option (b). Re-examining: AC + BD = (xy + yz + zx + z²)/z. The answer is (b).