In the given figure, \(\Delta ABC\) is a right-angled triangle with \(\angle A = 90^\circ\). AD is perpendicular to BC. 35(a)(i)Prove that \(\Delta DBA \sim \Delta DAC\)
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An important theorem states: "If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other."
Remembering this theorem helps you predict the similarity relations instantly.
Step 1: Understand the setup. $\Delta ABC$ is right-angled at $A$ (so $\angle BAC = 90^\circ$). $AD \perp BC$, so $\angle ADB = \angle ADC = 90^\circ$. We need to prove $\Delta DBA \sim \Delta DAC$. Step 2: Identify angles in $\Delta DBA$. In $\Delta DBA$: $\angle BDA = 90^\circ$ (AD is altitude). Let $\angle ABD = \angle B$. Then $\angle DAB = 90^\circ - \angle B$. Step 3: Identify angles in $\Delta DAC$. In $\Delta DAC$: $\angle ADC = 90^\circ$. $\angle ACD = \angle C$. Since in $\Delta ABC$, $\angle B + \angle C = 90^\circ$ (as $\angle A = 90^\circ$), we get $\angle DAC = 90^\circ - \angle C = \angle B$. Step 4: Apply AA similarity criterion. In $\Delta DBA$ and $\Delta DAC$: $\angle BDA = \angle ADC = 90^\circ$ and $\angle ABD = \angle DAC = \angle B$. By AA criterion: $\Delta DBA \sim \Delta DAC$. Proved! Step 5: State the corresponding sides ratio. From the similarity: \[ \frac{DB}{DA} = \frac{DA}{DC} = \frac{AB}{AC} \] This gives the important result: $DA^2 = DB \times DC$ (geometric mean relation). Step 6: Conclusion. By AA similarity criterion, $\Delta DBA \sim \Delta DAC$ is proved. \[ \boxed{\Delta DBA \sim \Delta DAC \text{ (proved by AA criterion)}} \]
