题目
如图所示,在△ABC中,AB=AC,点D在BC上,且CA=CD,DA=DB,求△ABC各角的度数
提问时间:2020-11-29
答案
∵CA=CD,∴∠CAD=∠CDA,∴∠ACB=180°-2∠CDA.
∵DA=DB,∴∠ABC=∠BAD,∴∠CDA=2∠ABC,∴2∠CDA=4∠ABC.
∴∠ACB=180°-4∠ABC.
∵AB=AC,∴ACB=∠ABC,∴∠ABC=180°-4∠ABC,∴5∠ABC=180°,∴∠ABC=36°,
∴∠ACB=36°,∴∠BAC=180°-2∠ABC=180°-2×36°=108°.
即△ABC中∠A、∠B、∠C的度数依次是:108°、36°、36°.
∵DA=DB,∴∠ABC=∠BAD,∴∠CDA=2∠ABC,∴2∠CDA=4∠ABC.
∴∠ACB=180°-4∠ABC.
∵AB=AC,∴ACB=∠ABC,∴∠ABC=180°-4∠ABC,∴5∠ABC=180°,∴∠ABC=36°,
∴∠ACB=36°,∴∠BAC=180°-2∠ABC=180°-2×36°=108°.
即△ABC中∠A、∠B、∠C的度数依次是:108°、36°、36°.
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