Question

如图,点C为线段AB上一点,△ACM,△CBN是等边三角形.

Answer 1

∵ AC = MC ,NC = BC ,∠MCB = ∠ACN = 120°

∴ △ACN ≌ △BCM

∴ AN = BM ,∠ANC = ∠CBM

∴ △CPN ≌ △CQB

∴ CP = CQ ,∠BCQ = ∠NCP

∵ ∠BCQ + ∠QCN = ∠BCN = 60°

∴ ∠NCP + ∠ QCN = ∠PCQ = 60°

∴ △CPQ为等边三角形(60°的等腰三角形)