Find: AD, AB.

Summary:

We will use trigonometric properties of right-angled triangles and the Pythagorean theorem to solve this problem.

Step-by-step solution:

1. Step 1: Analyze triangle BCD. It is a right-angled triangle with ∠BCD = 90°. We are given BC = 4. We are also given ∠CAD = 45°. In right triangle ABC, ∠BCA = 90°. In triangle ACD, ∠ADC = 90°. No. The diagram shows ∠BCA = 90°. And ∠ADC = 90°. So D is on AB. And CD is perpendicular to AB. In right triangle BCD, ∠BCD = 90°. This is incorrect, as C is a vertex of the triangle. Let's assume ∠BCA = 90°. D is a point on AB. And CD is perpendicular to AB. So ∠CDA = 90°. In triangle ABC, BC = 4. ∠BAC = 45°. So triangle ABC is a right isosceles triangle. This means BC = AC = 4. Then \( AB = sqrt{BC^2 + AC^2} = sqrt{4^2 + 4^2} = sqrt{16+16} = sqrt{32} = 4\]\[ ext{sqrt}(2) \). Since CD is the altitude to the hypotenuse, \( CD = rac{BC imes AC}{AB} = rac{4 imes 4}{4\]\[ ext{sqrt}(2)} = rac{4}{\]\[ ext{sqrt}(2)} = 2\]\[ ext{sqrt}(2) \). Also, \( AD = rac{AC^2}{AB} = rac{4^2}{4\]\[ ext{sqrt}(2)} = rac{16}{4\]\[ ext{sqrt}(2)} = rac{4}{\]\[ ext{sqrt}(2)} = 2\]\[ ext{sqrt}(2) \). And \( BD = rac{BC^2}{AB} = rac{4^2}{4\]\[ ext{sqrt}(2)} = rac{16}{4\]\[ ext{sqrt}(2)} = rac{4}{\]\[ ext{sqrt}(2)} = 2\]\[ ext{sqrt}(2) \). This assumes ∠BAC = 45°. However, the diagram shows ∠CAD = 45°. Let's re-examine the diagram. We have triangle ABC, with ∠BCA = 90°. D is a point on AB, and ∠CDA = 90°. So CD is the altitude to the hypotenuse AB. We are given BC = 4. We are given ∠CAD = 45°. So in right triangle ACD, ∠CAD = 45°, ∠CDA = 90°. This means triangle ACD is a right isosceles triangle. So CD = AD. Let CD = AD = x. Then \( AC = sqrt{AD^2 + CD^2} = sqrt{x^2 + x^2} = x\]\[ ext{sqrt}(2) \). Now consider right triangle BCD. ∠BDC = 90°. We have BC = 4. Also ∠CAD = 45°. In triangle ABC, ∠BCA = 90°. In triangle ACD, ∠CDA = 90°, ∠CAD = 45°. So ∠ACD = 45°. Thus, triangle ACD is isosceles with AD = CD. Let AD = CD = x. In right triangle BCD, we have BC = 4. We need to find BD. In triangle ABC, ∠BCA = 90°. Let ∠BAC = α. Then ∠ABC = 90° - α. In right triangle ACD, ∠CAD = 45°. So ∠ACD = 45°. Thus AD = CD. In right triangle BCD, ∠BDC = 90°. By Pythagorean theorem, \( BC^2 = BD^2 + CD^2 \). So \( 4^2 = BD^2 + x^2 \). \( 16 = BD^2 + x^2 \). In right triangle ACD, \( AC^2 = AD^2 + CD^2 = x^2 + x^2 = 2x^2 \). So \( AC = x\]\[ ext{sqrt}(2) \). In right triangle ABC, \( AB^2 = AC^2 + BC^2 = 2x^2 + 4^2 = 2x^2 + 16 \). Also, \( AB = AD + DB = x + BD \). So \( (x+BD)^2 = 2x^2 + 16 \). \( x^2 + 2x imes BD + BD^2 = 2x^2 + 16 \). Substitute \( BD^2 = 16 - x^2 \). \( x^2 + 2x imes BD + (16 - x^2) = 2x^2 + 16 \). \( 2x imes BD + 16 = 2x^2 + 16 \). \( 2x imes BD = 2x^2 \). Since x is a length, x ≠ 0. So \( BD = x \). Thus, BD = AD = CD = x. This means triangle BCD is also an isosceles right triangle, with ∠CBD = 45° and ∠BCD = 90°. This is consistent with ∠BCA = 90°. So triangle ABC is a right triangle with ∠BCA = 90°. CD is the altitude to AB. AD = CD = BD = x. Thus, \( BC^2 = BD^2 + CD^2 = x^2 + x^2 = 2x^2 \). So \( 4^2 = 2x^2 \). \( 16 = 2x^2 \). \( x^2 = 8 \). \( x = sqrt{8} = 2\]\[ ext{sqrt}(2) \). So AD = CD = BD = \( 2\]\[ ext{sqrt}(2) \). Then \( AB = AD + BD = 2\]\[ ext{sqrt}(2) + 2\]\[ ext{sqrt}(2) = 4\]\[ ext{sqrt}(2) \). And \( AC = x\]\[ ext{sqrt}(2) = 2\]\[ ext{sqrt}(2) imes \]\[ ext{sqrt}(2) = 2 imes 2 = 4 \). So AC = 4. This means triangle ABC is a right isosceles triangle with AC = BC = 4 and ∠BAC = ∠ABC = 45°. This contradicts ∠CAD = 45° given in the diagram, as D is on AB. Let's re-read the question and diagram carefully. We have a right triangle ABC, with ∠BCA = 90°. D is a point on AB such that ∠ADC = 90°. So CD is the altitude to the hypotenuse. We are given BC = 4. We are given ∠CAD = 45°. In right triangle ACD, ∠CDA = 90°, ∠CAD = 45°. Therefore, ∠ACD = 45°. So triangle ACD is an isosceles right triangle with AD = CD. Let AD = CD = x. In right triangle BCD, ∠BDC = 90°. We have BC = 4. By Pythagorean theorem, \( BC^2 = BD^2 + CD^2 \). \( 4^2 = BD^2 + x^2 \). \( 16 = BD^2 + x^2 \). In right triangle ABC, \( AB = AD + DB = x + BD \). Also \( AC = CD imes \]\[ ext{sqrt}(2) = x\]\[ ext{sqrt}(2) \). In right triangle ABC, \( AB^2 = AC^2 + BC^2 \). \( (x+BD)^2 = (x\]\[ ext{sqrt}(2))^2 + 4^2 = 2x^2 + 16 \). \( x^2 + 2x imes BD + BD^2 = 2x^2 + 16 \). Substitute \( BD^2 = 16 - x^2 \). \( x^2 + 2x imes BD + (16 - x^2) = 2x^2 + 16 \). \( 2x imes BD + 16 = 2x^2 + 16 \). \( 2x imes BD = 2x^2 \). Since x > 0, \( BD = x \). So AD = CD = BD = x. This implies triangle BCD is also an isosceles right triangle, with ∠CBD = 45°. Therefore, ∠ABC = 45°. Since ∠BCA = 90°, then ∠BAC = 45°. This means triangle ABC is an isosceles right triangle with AC = BC = 4. And AB = \( 4\]\[ ext{sqrt}(2) \). In this case, CD is the altitude to the hypotenuse. \( CD = rac{AC imes BC}{AB} = rac{4 imes 4}{4\]\[ ext{sqrt}(2)} = rac{4}{\]\[ ext{sqrt}(2)} = 2\]\[ ext{sqrt}(2) \). Also \( AD = rac{AC^2}{AB} = rac{4^2}{4\]\[ ext{sqrt}(2)} = rac{16}{4\]\[ ext{sqrt}(2)} = rac{4}{\]\[ ext{sqrt}(2)} = 2\]\[ ext{sqrt}(2) \). And \( BD = rac{BC^2}{AB} = rac{4^2}{4\]\[ ext{sqrt}(2)} = rac{16}{4\]\[ ext{sqrt}(2)} = rac{4}{\]\[ ext{sqrt}(2)} = 2\]\[ ext{sqrt}(2) \). So AD = \( 2\]\[ ext{sqrt}(2) \). This is consistent. Let's check the given angle ∠CAD = 45°. If triangle ABC is isosceles right with AC = BC = 4, then ∠BAC = 45°. Since D is on AB, and CD is the altitude, then AD = \( 2\]\[ ext{sqrt}(2) \). So ∠CAD = ∠BAC = 45°. This is consistent. Therefore, AD = \( 2\]\[ ext{sqrt}(2) \).
2. Step 2: Find AB. \( AB = AD + BD = 2\]\[ ext{sqrt}(2) + 2\]\[ ext{sqrt}(2) = 4\]\[ ext{sqrt}(2) \).

Answer: AD = 2√2, AB = 4√2