In the triangle ABC, the angle ABC = 45 degrees, ad vertical BC. The perpendicular foot is d. be vertical AC. the perpendicular foot is e. ad intersects be with F, connecting cf. if the angle BAC is an acute angle In the triangle ABC, the angle ABC = 45 degrees, ad is vertical BC. The perpendicular foot is d. be vertical AC. the perpendicular foot is e. ad intersects be with F and connects cf. if the angle BAC is an acute angle, it is proved that the triangle CDF is an isosceles right triangle

In the triangle ABC, the angle ABC = 45 degrees, ad vertical BC. The perpendicular foot is d. be vertical AC. the perpendicular foot is e. ad intersects be with F, connecting cf. if the angle BAC is an acute angle In the triangle ABC, the angle ABC = 45 degrees, ad is vertical BC. The perpendicular foot is d. be vertical AC. the perpendicular foot is e. ad intersects be with F and connects cf. if the angle BAC is an acute angle, it is proved that the triangle CDF is an isosceles right triangle

Angle AFE = angle BFD (opposite vertex angle)
The AFE of RT triangle is similar to BFD of RT triangle
Angle CAD = angle FBD
RT triangle in abd
Angle ABC = 45 degrees
So ad = BD
RT triangle ACD and RT triangle BFD congruence
FD=CD
Ad vertical BC
So the triangle CDF is a right triangle

In the known triangle ABC, the angle ACB is 90 degrees, the angle CBA is 45 degrees, e is a point on AC, extend BC to point D, make CD = CB

The intersection of be and ad should be "CD = CE" when f ∫ ACB = 90 °, CBA = 45 °, ACB is an isosceles right triangle, AC = BC ∫ ACD = ∠ ACB = ∠ BCE = 90 °, ACD and △ BCE are right triangles, in RT △ ACD and RT △ BCE, AC = BCCD = CE ≌ RT △ ACD ≌ RT △ BCE ≌ DAC = ∠ EBC, that is ≌ FAE

In △ ABC, ∠ BAC is an acute angle, ab = AC, ad and be are high, they intersect at point h, and AE = be. (1) it is proved that ah = 2bd; (2) if ∠ BAC is changed to an obtuse angle, and other conditions remain unchanged, the above conclusion is still true? If yes, please prove; if not, please give reasons

(1) As shown in Figure 1, the figure 1, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\therest As shown in Fig. 2, ∵ ad ⊥ BC, be ⊥ AE, ∵ CAD + ⊥ C = 90 °, ≌ CBE + ≌ C = 90 ° and ≌ CAD = ≌ CBE, i.e. ≁ eah = ≁ CBE, in △ ahe and △ BCE, ≁ eah = ≁ cbeae = be ≌ AEH = ≌ BEC = 90 °, ah = BC, ≌ AB = AC, ad ⊥ BC, ≁ BD = CD = 12bc, i.e., BC = 2bd, then ah = 2bd

In △ ABC, ∠ BAC is an acute angle, ab = AC, ad and be are high, they intersect at point h, and AE = be. (1) it is proved that ah = 2bd; (2) if ∠ BAC is changed to an obtuse angle, and other conditions remain unchanged, the above conclusion is still true? If yes, please prove; if not, please give reasons

(1) It is proved that: as shown in Fig. 1, ∵ ad ⊥ BC, be ⊥ AC, ∵ CAD + C = 90 °, CBE + C = 90 ° and ≌ CAD = CBE, that is, ∵ eah = CBE, in △ ahe and △ BCE, ≌ eah = cbeae = be ≌ AEH = BEC = 90 ° and ≌ aha ≌ BCE (ASA), ≌ ah = BC, ≌ AB = AC, ad