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.
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I. Rectangle
II. Isosceles triangle
