In △ ABC, if high AD and be intersect at H point and BH = AC, then ∠ ABC=______ .

There are two cases, as shown in figures (1), (2),
∵ Bhd = ∠ ahe, and ∠ AEH = ∠ ADC = 90 °,
∴∠DAC+∠C=90°，∠HAE+∠AHE=90°，
∴∠AHE=∠C，
∴∠C=∠BHD，
∵BH=AC，∠HBD=∠DAC，∠C=∠BHD，
∴△HBD≌△CAD，
∴AD=BD．
As shown in figure (1), ∠ ABC = 45 °;
As shown in Fig
∵AD=BD，AD⊥BD，
△ ADB is an isosceles right triangle,
∴∠ABD=45°，
∴∠ABC=180°-45°=135°，
So the answer is: 45 ° or 135 °

As shown in the figure, in trapezoid ABCD, ad ∥ BC, ∠ ABC = 90 °, DG ⊥ BC on G, BH ⊥ DC on H, ch = DH, point E on AB, point F on BC, and ef ∥ DC （1） If AD=3, CG=2, find CD; (2) If CF = AD + BF, verification: EF = 1 2CD．

(1) Even BD, as shown in the figure, ∵ in trapezoid ABCD, ad ∵ BC, ∵ ABC = 90 °, DG ⊥ BC, ? abgd is a rectangle, ∵ ad = BG = 3, ab = DG, and ∵ BH ⊥ DC, ch = DH, ? BDC is an isosceles triangle, ? BD = BG + GC = 3 + 2 = 5, in RT △ abd, ab = BD2 − ad2 = 52 − 32 = 4, ? DG = 4

As shown in the figure, in △ ABC, h is high, h is the intersection point of high AD and be, ad = BD, verification: DH = DC

A right triangle ADC and a right triangle BEC share a common angle c, so the angle CAE is equal to the angle EBD;
And because ad = BD,
So right triangle HBD and right triangle CAD are congruent
So HD = DC

As shown in the figure, in the equilateral △ ABC, points D and E are on sides BC and ab respectively, and BD = AE, ad and CE intersect at point F, then the degree of ∠ DFC is () A. 60° B. 45° C. 40° D. 30°

∵ △ ABC is an equilateral triangle
∴∠BAC=∠ABC=∠BCA=60°
∴AB=BC=AC
In △ abd and △ CAE
BD=AE，∠ABD=∠CAE，AB=AC
∴△ABD≌△CAE
∴∠BAD=∠ACE
And ? bad +  DAC =  BAC = 60 
∴∠ACE+∠DAC=60°
∵∠ACE+∠DAC+∠AFC=180°
∴∠AFC=120°
∵∠AFC+∠DFC=180°
∴∠DFC=60°．
Therefore, a

As shown in the figure, in △ ABC, ∠ B = 22.5 °, the vertical bisector of edge AB intersects BC at D, DF ⊥ AC at F, and intersects with high AE on BC edge at g. verification: eg = EC

prove:
Connect ad,
∵ the vertical bisector of side AB intersects BC at D,
∴BD=AD，
∴∠B=∠BAD=22.5°，
∴∠ADE=22.5°+22.5°=45°，
∵AE⊥BC，
∴∠AEC=∠AED=90°，
∴∠DAE=45°=∠ADE，
∴DE=AE，
∵DF⊥AC，
∴∠DFC=90°=∠AEC，
∴∠ACE+∠FDC=90°，∠ACD+∠CAE=90°，
∴∠CAE=∠FDC，
In △ DEG and △ AEC
∠DEA＝∠AEC
DE＝AE
∠GDE＝∠CAE
∴△DEG≌△AEC（ASA），
∴EG=EC．

As shown in the figure, in △ ABC, ∠ B = 22.5 °, the vertical bisector of edge AB intersects BC at D, DF ⊥ AC at F, and intersects with high AE on BC edge at g. verification: eg = EC

In the isosceles triangle ABC, ab = AC, ad ⊥ BC at point D, CG parallel to AB, BG intersecting ad respectively, AC at point E, F

Connect CE
∵AB=AC
∴∠B=∠C
∵AD⊥BC
﹤ BD = CD (the vertical line of the bottom edge of an isosceles triangle is the center line and the angular bisector)
The ad is the vertical bisector of BC
Therefore, be = CE
In △ Abe and △ ace
∵AB=AC,BE=CE,AE=AE
∴△ABE≌△ACE
∴∠ABE=∠ACE
∵CG‖AB
∴∠ABE=∠CGE
∵∠ABE=∠ACE
∴∠ACE=∠CGE
In △ CEF and △ CEG
∵∠FEC=∠GEC,∠FCE=∠CGE
∴△CEF∽△CEG
∴CE/EG=EF/CE,CE^2=EF×EG
∵CE=BE
∴BE^2=EF×EG

In the isosceles triangle ABC, ab = AC, ad is perpendicular to D, CG ∥ AB, BG intersect ad respectively, AC is in E, F. it is proved that be × be = EF × eg

If EC is connected, be = CE
CE * ce = EF * eg
As long as it is proved that the triangle EFC is similar to the triangle ECG (the three internal angles are equal)
Thus EF / EC = EC / eg
Proof of the original proposition

In △ ABC, if high AD and be intersect at H point and BH = AC, then ∠ ABC=______ .