As shown in the figure, given that AD and be are the heights of △ ABC, ad and be intersect at point F, and ad = BD, can you find the congruent triangle in the figure? If you can find it, please explain why

As shown in the figure, given that AD and be are the heights of △ ABC, ad and be intersect at point F, and ad = BD, can you find the congruent triangle in the figure? If you can find it, please explain why


The reasons are as follows: ≔ ad ⊥ BC, be ⊥ AC, ∽ ADC = ≌ BEC = 90 ° and ≌ CBE = ≌ CAD, in ≌ ADC and ≌ BDF, ≌ CAD = ≌ cbead = BD ⊥ ADC = ≌ BDF, ≌ ADC ≌ BDF (ASA)



Verification: as shown in the figure, the corresponding midlines of congruent triangles are equal. It is known that △ ABC ≌ △ a'b'c 'and ad a'd' are the midlines of △ ABC and a'b'c 'respectively=


∵ΔABC≌ΔA'B'C'
∴∠B=∠B',BC=B'C',AB=A'B'
BD = 1 / 2BC, b'd '= 1 / 2b'c'
∴BD=B'D'
In Δ abd and Δ a'b'd '
AB=A'B',∠B=∠B',BD=B'D'
∴ΔABD≌ΔA'B'D'(SAS)
∴AD=A'D



As shown in the figure, the known points E and C are on the line BF, be = CF, ab ‖ De, ∠ ACB = ∠ F


It is proved that: ∵ ab ‖ De, ∵ B = ∵ def. ∵ be = CF, ∵ BC = EF. ∵ ACB = ∵ F, ∵ B = ≌ defbc = EF ≌ ACB = ≌ F, ≌ ABC ≌ def (ASA)



As shown in the figure, the known points E and C are on the line BF, be = CF, ab ‖ De, ∠ ACB = ∠ F


It is proved that: ∵ ab ‖ De, ∵ B = ∵ def. ∵ be = CF, ∵ BC = EF. ∵ ACB = ∵ F, ∵ B = ≌ defbc = EF ≌ ACB = ≌ F, ≌ ABC ≌ def (ASA)