$解:(1)∵AB=AC=BD$
$∴∠ABC=∠C=50°,∠BAC=∠ADB=180°-2∠C=80°$
$∴∠ABD=180°-∠A-∠ADB=20°$
$(2)过A作AM⊥BC于M,过B作BN⊥AC于N$
$∵AM是等腰三角形ABC底边上的高$
$∴AM是BC中线,即BM=CM=3$
$∴AM=\sqrt {AB^{2}-BM^{2}}=4$
$∵S_{△ABC}=\frac {1}{2}BC×AM=\frac {1}{2}AC×BN$
$∴解得BN=\frac {24}{5}$
$∴可求得AN=\sqrt {AB^{2}-BN^{2}}=\frac {7}{5}$
$∴DN=AN=\frac {7}{5}$
$AD=AN+DN=\frac {14}{5}$
