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Ask for help with the ninth grade math similar triangles problem ~
AB = ABCD in parallelogram

∠EDC+∠C=90 degrees, ∠FBC+∠C=90 degrees.

∴∠EDC=∠FBC

∠∠DEC =∠DEB = 90,BE=DE

∴⊿BHE≌⊿DCE

∴BH=DC

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In parallelogram, ABCD ∠A=∠C, AD ∠ BC.

∴∠G=∠GBC

∠∠c =∠bhe (obtained by ⊿BHE≌⊿DCE)

∴∠A=∠BHE

∴⊿ABG∽⊿HEB(∠G=∠GBC,∠A=∠BHE)

∴AB:AG=HE:BH

∴AB*BH=AG*AE

AB = BH

∴AB square =GA*HE