https://bgk-photo.cdn.bcebos.com/doc/pic/item/08f790529822720e88af64d073cb0a46f21fabb3.jpg如图，已知AD，BE，CF分别是△ABC三边的高，H是垂心，AD的延长线交△ABC的外接圆于点G．求证：DH=DG．
https://bgk-photo.cdn.bcebos.com/doc/pic/item/08f790529822720e88af64d073cb0a46f21fabb3.jpg如图，已知AD，BE，CF分别是△ABC三边的高，H是垂心，AD的延长线交△ABC的外接圆于点G．求证：DH=DG．
发布时间：2025-02-25 16:45:59
解：连结CG，∵AD⊥BC，∴∠ABC ∠GAB=90°同理可得∠ABC ∠FCB=90°，从而得到∠GAB=∠FCB=90°-∠ABC又∵∠GAB与∠GCB同对弧BG，∴∠GAB=∠GCB，可得∠GCB=∠FCB，∵CD⊥GH，即CD是△GCH的高线∴△CHG是以HG为底边的等腰三角形，可得DH=DG．
