△DBC≡△EACより∠AEC=∠BDC
∠ACD＋∠AEC=∠ACD＋∠BDC
∠ACDも∠BDCも△ADCの内角だから、その和は外角∠BACに等しい
よって60°
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△DBC≡△EACより∠AEC=∠BDC
∠ACD＋∠AEC=∠ACD＋∠BDC
∠ACDも∠BDCも△ADCの内角だから、その和は外角∠BACに等しい
よって60°
Users viewing this question
are also looking at these questions 😉
ありがとうございます!!わかりました✨