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)