In the given figure, \(\triangle\)ABC is right angled triangle with \(\angle\)A = \(90^\circ\). AD is perpendicular to BC. (a)(i)Prove that \(\triangle\)DBA \(\sim\) \(\triangle\)DAC
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A useful theorem to remember is: "The perpendicular drawn from the vertex of the right angle of a right triangle to the hypotenuse divides the triangle into two triangles which are similar to each other and to the original triangle."
Step 1: Understand the figure. Triangle ABC is right-angled at A. AD is perpendicular to BC. We need to prove \(\triangle DBA \sim \triangle DAC\). Step 2: Identify angles in triangle DBA. In \(\triangle DBA\): \(\angle ADB = 90^\circ\) (AD perpendicular to BC). Let \(\angle ABD = \angle B\). Then \(\angle DAB = 90^\circ - \angle B\). Step 3: Identify angles in triangle DAC. In \(\triangle DAC\): \(\angle ADC = 90^\circ\), and \(\angle ACD = \angle C\). So \(\angle DAC = 90^\circ - \angle C\). Step 4: Use the angle sum in triangle ABC. In \(\triangle ABC\): \(\angle B + \angle C = 90^\circ\) (since \(\angle A = 90^\circ\)), meaning \(\angle C = 90^\circ - \angle B\). So \(\angle DAB = 90^\circ - \angle B = \angle C = \angle ACD\). Step 5: Match corresponding angles. \(\angle ADB = \angle ADC = 90^\circ\) and \(\angle ABD = \angle DAC\). So by AA criterion: \(\triangle DBA \sim \triangle DAC\). Step 6: State the conclusion. By the AA (Angle-Angle) Similarity Criterion, \(\triangle DBA \sim \triangle DAC\) is proved. \[ \boxed{\triangle DBA \sim \triangle DAC \text{ (by AA criterion)}} \]
