Solution: extend AD to E to make AD=DE, and connect be≈ADC =∠b dead = debd = DC ∴△ ADC △ EDB ∴ Be = AC = 3 ∫ AB = 5ae = 2ad = 2×.
According to Pythagorean theorem AB squared AC squared =BC squared, and according to triangle area AD multiplied by BC=AC multiplied by AB, the equations AC=3 and BC=5 are solved.