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20 10 national mathematics 12
14. solution: (1) ∵AE⊥BD, ∴BE⌒=DE⌒, ∴ EBD = ∠ ECB. ∠∠ABH。

∠ DBH +∠EBD, ∴∠ BHE = ∠ HBE. Yes = him ...............................................................................................................

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