As shown in the figure, △ ABC, Ag ⊥ BC is at point G, AB and AC are taken as one side to make rectangle ABME and rectangle acnf outside △ ABC respectively, and the ray GA intersects EF at point h. If AB = KAE, AC = Kaf, try to explore the quantitative relationship between he and HF, and explain the reason

He = HF. Reason: EP ⊥ GA and FQ ⊥ GA are made through point E, and the perpendicular feet are p and q. ∵ quadrilateral. ABME is rectangular, ∵ BAE = 90 °, and ∵ Ag ⊥ BC, ∵ bag + ABG = 90 °, and ∵ ABG = ∵ EAP. ∵ AGB = ∵ EPA = 90 °, and ∵ ABG ∽ EAP, ∵ Ag: EP = AB: EA In RT △ Eph and RT △ FQH, ∠ Eph = ∠ FQA ∠ EHP = ∠ fhqep = FQ,  RT △ Eph ≌ RT △ FQH (AAS).. he = HF

In the triangle ABC, ab = AC, if BC = BD = de = EF = FG = GA

Angle a equals 180 / 11

An ant starts from a, crawls along the polygon for a circle and returns to point a, then vector AB + vector BC + vector CD + vector de + vector EF + vector FG + vector GA = ()

Is a zero vector, only with the starting point, the end point, has nothing to do with the process

As shown in the figure, if ∠ a = 15 ° AB = BC = CD = de = EF, then ∠ DEF is equal to______ ．

∵AB=BC=CD=DE=EF，∠A=15°，∴∠BCA=∠A=15°，∴∠CBD=∠BDC=∠BCA+∠A=15°+15°=30°，∴∠BCD=180°-（∠CBD+∠BDC）=180°-60°=120°，∴∠ECD=∠CED=180°-∠BCD-∠BCA=180°-120°-15°=45°，∴∠CDE=180°-（∠ECD+∠CED）=180°-90°=90°，∴∠EDF=∠EFD=180°-∠CDE-∠BDC=180°-90°-30°=60°，∴∠DEF=180°- (EDF + EFD) = 180 ° - 120 ° = 60 °. So the answer is: 60 °

As shown in the figure, the line AB and EF intersect at point E, the line CD and EF intersect at point F, the eg bisection angle bef, the FG bisection angle EFD, and the EG is vertical to FG. Verify that AB is parallel to CD
Right now
_______ E_ #_ B
A # \
#There's another line in the middle of G
C #____ D is a right triangle in the middle. Let's make do with it
F

Where is the picture?
It's so humorous
.
prove:
Because eg bisector bef, FG bisector EFD
And eg vertical FG
So the angle bef + the angle EFD = 180
So AB parallels CD
The inner angles of the same side complement each other, and the two lines are parallel

As shown in the figure, the straight line EF intersects AB at e, intersects CD at F, eg bisects ∠ AEF, FG bisects ∠ EFC, they intersect at g, if ∠ EGF = 90 ° verify: ab ‖ CD

∵ eg bisection ∠ AEF, FG bisection ∠ EFC
∴∠GEF=∠GEA=1/2∠AEF
∠GFC=∠GFE=1/2∠EFC
∵∠EGF=90°
∴∠GEF+∠GFE=180°-∠EGF=180°-90°=90°
∴∠AEF+∠EFC=2（∠GEF+∠GFE）=180°
∴AB∥CD

23. As shown in the figure, ab ∥ CD is known. The straight line EF intersects AB and CD at points E and f respectively. Eg bisects ∠ bef. If ∠ 1 = 50 °, then ∠ 2 is ()

What about the picture?

It is known that as shown in Figure 4, the line ab ‖ CD and the line EF intersect AB respectively
CD intersects with E, bisector of ∠ bef intersects with bisector of ∠ DFE at point P, try to find the size of ∠ P

∵AB∥CD
∴∠BEF+∠DFE=180°
And ∵ PE bisection ∠ bef
Pf bisection ∠ DFE
∴∠PEF=1/2∠BEF
∠PFE=1/2∠DFE
∴∠PEF+∠PFE =1/2(∠∠BEF+∠DFE)=90°
In ∵ triangle PEF, ∠ P = 180 ° - (∠ PEF + ∠ PFE)
∴∠P=90°

It is known that in △ ABC, ∠ ABC = ∠ ACB, D moves on the ray Ca (not coincident with a and C), and takes C as the vertex, AC as the side to make ∠ ACP = ∠ cbdpc intersect with ray dB
The intersection point of PC and DB is p
(1) If point d moves on line AC
① If ∠ BAC = 40 °, calculate the degree of ∠ BPC
② If ∠ BAC = n °, find the degree of ∠ BPC (expressed by an algebraic expression containing n)

It is known that in △ ABC, ∠ ABC = ∠ ACB, D moves on the ray Ca (not coincident with a and C). Take C as the vertex, AC as one side, make ∠ ACP = ∠ CBD, PC and ray DB intersect at point P. if point d moves on the line AC, (1) if ∠ BAC = 40 °, calculate the degree of ∠ BPC; (2) if ∠ BAC = n °, calculate the degree of ∠ BPC (expressed by an algebraic formula containing n)
(1) Analysis: ∵ in △ ABC, ∵ ABC = ∵ ACB, ∵ AB = AC
∵∠BAC=40°,∴∠ABC=∠ACB=(180-40)/2=70°
In ⊿ BDC and ⊿ CDP
∵∠ACP=∠CBD,∠PDC=∠CDB
∴⊿BDC∽⊿CDP==>∠DPC=∠DCB=70°
∴∠BPC=110°
(2) Analysis: ∵ ∠ BAC = n °
According to (1), BPC = 180 - (180-n) / 2 = ((180 + n) / 2 degrees
It can be seen that the size of BPC does not change during the motion of point D on line AC

As shown in the figure, it is known that the diamond AMNP is inscribed with △ ABC, and m, N and P are respectively on AB, BC and ac. if AB = 21cm and Ca = 15cm, the perimeter of the diamond AMNP is calculated

∵ AMNP is rhombic, ∥ PN ∥ AB, ∥ CPN ∥ cab, ∥ CP: CA = PN: AB, ∵ PN = PA, ∥ CP: CA = PA: AB, that is, CP: 15 = PA: 21, ∥ CP: PA = 15:21 = 5:7, ∥ CP + PA: Pa = (5 + 7): 7, ∥ AC: PA = 12:7, that is, 15: PA = 12:7, the solution is pa = 354 (CM), and the perimeter of ∥ rhombic AMNP is 354 × 4 = 35cm