1. **Stating the problem:** We have a right-angled triangle $\triangle ABC$ with the right angle at vertex $A$. The line segment $AD$ is perpendicular to $BC$. We need to find which of the given statements is false: (a) $\triangle ABC \sim \triangle DBA$ (b) $\triangle ABC \sim \triangle DAC$ (c) $\triangle ACD \sim \triangle BAD$ (d) $AD = DB \times DC$ 2. **Using properties of right triangles and similarity:** Since $\triangle ABC$ has $\angle A = 90^\circ$ and $AD \perp BC$, by altitude-on-hypotenuse theorem, the altitude $AD$ divides $\triangle ABC$ into two smaller right triangles $\triangle ADB$ and $\triangle ADC$. 3. **Checking similarity:** - $\triangle ABC$ and $\triangle DBA$ share $\angle B$. Both have a right angle: $\angle A = 90^\circ$ and $\angle D = 90^\circ$. Therefore, by AA (Angle-Angle) similarity, $\triangle ABC \sim \triangle DBA$ (statement a) is true. - Similarly, $\triangle ABC$ and $\triangle DAC$ share $\angle C$ and both have a right angle. Thus $\triangle ABC \sim \triangle DAC$ (statement b) is true. - For $\triangle ACD$ and $\triangle BAD$, they share no common angles explicitly. However, since $AD$ is perpendicular to $BC$, both triangles are right angled at $D$. The angles $\angle ACD = \angle BAD$ (alternate angles) if $AD$ is altitude, so $\triangle ACD \sim \triangle BAD$ (statement c) is true. 4. **Checking the segment relation:** The altitude length $AD$ in a right triangle relates to segments $DB$ and $DC$ on the hypotenuse by the formula: $$AD^2 = DB \times DC$$ This means: $$AD = \sqrt{DB \times DC}$$ Therefore, statement (d) $AD = DB \times DC$ is false. **Final answer: (d) $AD = DB \times DC$ is the false statement.**