1. **Problem Statement:** Given triangle ABC, right-angled at B, and D is the midpoint of AC. We need to prove that $DA = DB = DC$. 2. **Understanding the problem:** Since $D$ is the midpoint of $AC$, it divides $AC$ into two equal parts. Also, $\angle B$ is $90^\circ$, so triangle $ABC$ is a right triangle. 3. **Given:** - $\triangle ABC$ with $\angle B = 90^\circ$ - $D$ is midpoint of $AC$ so $AD = DC$ 4. **To prove:** $DA = DB = DC$ 5. **Proof:** - Since $D$ is midpoint of $AC$, $AD = DC$ (by definition of midpoint). - We need to show $DB = DA$. 6. **Using the midpoint theorem and properties of right triangles:** - In right triangle $ABC$, $AC$ is the hypotenuse. - Point $D$ is midpoint of hypotenuse $AC$. - A key property: The midpoint of the hypotenuse of a right triangle is equidistant from all three vertices. 7. **Therefore:** - $DB = DA = DC$ 8. **Explanation:** - Since $D$ is midpoint of hypotenuse $AC$, $D$ lies on the circle with diameter $AC$. - This circle passes through $B$ because $\angle B$ is right angle. - Hence, $DB = DA = DC$. **Final answer:** $DA = DB = DC$