三角形ABC的三條角平分線AD,BE,CF相交于點(diǎn)O,則角EBC+角DOC=( )度
三角形ABC的三條角平分線AD,BE,CF相交于點(diǎn)O,則角EBC+角DOC=( )度
數(shù)學(xué)人氣:209 ℃時(shí)間:2020-06-21 03:31:12
優(yōu)質(zhì)解答
應(yīng)該說角EBC+角DOC=90度
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