Prove that AB² = AD * AC.

Solution:

• The problem asks to prove $$AB^2 = AD \cdot AC$$.
• This suggests we need to find similar triangles where the sides are in proportion to AB, AD, and AC.
• Consider triangles ABC and ABD.
• Angle A is common to both triangles.
• We need another pair of equal angles.
• The angle between the tangent AB and chord BC is $$\angle ABC$$. This angle subtends arc BC. The inscribed angle subtended by arc BC is $$\angle BAC$$. Therefore, $$\angle ABC = \angle BAC$$. This is incorrect as $$\angle BAC$$ is not related to arc BC in the standard way. The angle between tangent AB and chord BC is equal to the inscribed angle subtended by arc BC, which is $$\angle BDC$$. So, $$\angle ABC = \angle BDC$$. This is also incorrect based on the diagram.
• Let's consider triangles ABC and ADB.
• Angle A is common to both.
• Angle ABC is subtended by arc AC. Angle ADC is also subtended by arc AC. So, $$\angle ABC = \angle ADC$$.
• Therefore, by AA similarity, $$\triangle ABC \sim \triangle ADB$$.
• This gives the proportion: $$\frac{AB}{AD} = \frac{AC}{AB} = \frac{BC}{DB}$$.
• From $$\frac{AB}{AD} = \frac{AC}{AB}$$, cross-multiplying gives $$AB \cdot AB = AD \cdot AC$$, which is $$AB^2 = AD \cdot AC$$.

Answer: Triangles ABC and ADB are similar by AA similarity because $$\angle A$$ is common to both triangles and $$\angle ABC = \angle ADC$$ (angles subtended by the same arc AC). This similarity leads to the proportion $$\frac{AB}{AD} = \frac{AC}{AB}$$, which rearranges to $$AB^2 = AD \cdot AC$$.