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Math problems about triangles in grade three.
Proof: (1) ∠ BDC = ∠ BEC = ∠ CDA = 90, ∠ ABC = 45,

∴∠BCD=45 =∠ABC,∠A+∠DCA=90,∠A+∠ABE=90,

∴DB=DC,∠ABE=∠DCA,

∫In△DBH and △DCA

∠BDH=∠CDA BD=CD ∠HBD=∠ACD,

∴△DBH≌△DCA,

∴BH=AC.

(2) connecting CG,

F is the midpoint of BC, DB=DC,

∴DF vertically divides BC,

∴BG=CG,

∵∠ABE=∠CBE,BE⊥AC,

∴∠AEB=∠CEB,

In △ABE and △CBE.

∠∠ AEB =∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠,

∴△ABE≌△CBE,

∴EC=EA,

In Rt△CGE, the square of BG-the square of GE = the square of EA is obtained by Pythagorean theorem.