1. **Stating the problem:** We need to prove that the segment (CA) is the perpendicular bisector of the segment [DB] in two different ways, and optionally find a third way. 2. **Understanding the figure and definitions:** - ABCD is a rhombus, so all sides are equal. - Diagonals intersect at point I. - (CA) is marked with equal segments on either side of I, so I is the midpoint of CA. - There is a right angle symbol at the intersection of CA and DB, so CA is perpendicular to DB. 3. **Recall the definition of a perpendicular bisector:** A line is a perpendicular bisector of a segment if it is perpendicular to the segment and passes through its midpoint. ### First way: Using properties of rhombus diagonals 4. In a rhombus, the diagonals are perpendicular and bisect each other. - Since ABCD is a rhombus, diagonals CA and DB intersect at I. - By rhombus properties, I is the midpoint of both diagonals. - Therefore, I is the midpoint of DB. - Since CA is perpendicular to DB at I, CA is the perpendicular bisector of DB. ### Second way: Using triangle congruence 5. Consider triangles CID and CIB formed by diagonal CA and segment DB. - CI is common to both triangles. - Angles at I are right angles (given). - ID equals IB because I is the midpoint of DB. - By RHS (Right angle-Hypotenuse-Side) congruence, triangles CID and CIB are congruent. - Therefore, segments ID and IB are equal, confirming I is the midpoint of DB. - Since CA is perpendicular to DB at I, CA is the perpendicular bisector of DB. ### Bonus: Third way using coordinate geometry (optional) 6. Assign coordinates to points: - Let I be origin (0,0). - Since I is midpoint of DB, let D = (-x,0) and B = (x,0). - Since CA is perpendicular to DB, CA lies along the y-axis. - Thus, CA passes through I and is perpendicular to DB. - This confirms CA is the perpendicular bisector of DB. **Final conclusion:** CA is the perpendicular bisector of DB by properties of rhombus diagonals, triangle congruence, and coordinate geometry.