In the given figure, \(\Delta ABC\) is a right-angled triangle with \(\angle A = 90^\circ\). AD is perpendicular to BC. 34(a)(i)Prove that \(\Delta DBA \sim \Delta DAC\)
Show Hint
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: State the Given Information. In $\Delta ABC$, $\angle BAC = 90^\circ$ (right angle at $A$). $AD$ is drawn perpendicular to $BC$, so $\angle ADB = \angle ADC = 90^\circ$. We need to prove $\Delta DBA \sim \Delta DAC$. Step 2: Identify the Angles in $\Delta DBA$. In $\Delta DBA$: $\angle ADB = 90^\circ$ and let $\angle DBA = \angle B$. So the third angle $\angle DAB = 90^\circ - \angle B$. Step 3: Identify the Angles in $\Delta DAC$. In $\Delta DAC$: $\angle ADC = 90^\circ$ and let $\angle DCA = \angle C$. The third angle $\angle DAC = 90^\circ - \angle C$. Step 4: Use the Angle Sum in the Original Triangle. In $\Delta ABC$: $\angle A + \angle B + \angle C = 180^\circ$. Since $\angle A = 90^\circ$, we have $\angle B + \angle C = 90^\circ$, so $\angle C = 90^\circ - \angle B$. Step 5: Compare the Angles of the Two Triangles. In $\Delta DBA$: angles are $90^\circ$, $\angle B$, and $90^\circ - \angle B$. In $\Delta DAC$: angles are $90^\circ$, $\angle C = 90^\circ - \angle B$, and $\angle DAC = \angle B$. So both triangles have the same set of angles: $90^\circ$, $\angle B$, and $90^\circ - \angle B$. By the AA criterion, $\Delta DBA \sim \Delta DAC$. Step 6: Write the Correspondence and Conclude. The vertices correspond as: $D \leftrightarrow D$, $B \leftrightarrow A$, $A \leftrightarrow C$. Therefore, by AA similarity: \[ \boxed{\Delta DBA \sim \Delta DAC \text{ (by AA similarity criterion)}} \]
