Geometric Configuration and Cyclic Property

The problem involves two right-angled triangles, \(\triangle ABC\) with \(\angle BAC = 90^\circ\) and \(\triangle DBC\) with \(\angle BDC = 90^\circ\). Both triangles share the hypotenuse BC. Since the angles \(\angle BAC\) and \(\angle BDC\) subtend the same segment BC and are both right angles (\(90^\circ\)), the points A and D must lie on the circle having BC as its diameter. Therefore, ABDC forms a cyclic quadrilateral.

Intersecting Diagonals Theorem Application

The diagonals of the cyclic quadrilateral ABDC are AC and DB, which intersect at point P. For a cyclic quadrilateral, the product of the segments of the intersecting diagonals is equal. This is given by the theorem:

\(AP \cdot PC = BP \cdot PD\)

Derivation of Length PD

Given the lengths:

• \(AP = x\)
• \(PC = y\)
• \(BP = z\)

Substituting these values into the intersecting diagonals theorem:

\(x \cdot y = z \cdot PD\)

Solving for PD yields:

\(PD = \frac{xy}{z}\)

Lengths AC and BD Calculation

The lengths of the sides AC and BD can be calculated as follows:

• \(AC = AP + PC = x + y\)
• \(BD = BP + PD = z + \frac{xy}{z}\)

Summation AC + BD

The required sum is \((AC + BD)\):

\(AC + BD = (x+y) + \left(z + \frac{xy}{z}\right)\)

\(AC + BD = x + y + z + \frac{xy}{z}\)

To express this result in a form matching the options, we find a common denominator, \(z\):

\(AC + BD = \frac{xz}{z} + \frac{yz}{z} + \frac{z^2}{z} + \frac{xy}{z}\)

\(AC + BD = \frac{xy + yz + zx + z^2}{z}\)

This final expression matches Option B.