Question

如图 已知BC=AC=AD,∠BAD=69°,求∠CAD的度数

Answer 1

∠CAD=180º-2∠ACD=180-4∠BAC
两边减去4∠CAD得
-3∠CAD=180º-4∠BAC-4∠CAD=180º-4*69º=-96º
∠CAD=32º

Answer 2

∵BC=AC=AD
∴∠B=∠BAC , ∠ACD=∠D
∵∠BAC+∠B=∠ACD
∴2∠BAC=∠ACD
∵∠CAD =180°-2∠ACD
∴∠CAD=180°-4∠BAC
∵∠BAD=∠BAC+∠CAD=69°
∴∠BAC+180°-4∠BAC=69°
∠BAC=37°
∴∠CAD=69°-37°
=32°

Answer 3

证明：假设∠B=x；∠D=y
因为BC=AC=AD
所以∠B=∠BAC=x；∠D=∠CAD=y
又∠ACD=∠B+∠BAC；即y=2x
因为在三角形ACD中180°=2y+69°-x
而y=2x；代入上式得x=37°
故∠CAD=32°

Answer 4

32