As shown in the figure, in the triangle ABC, the vertical bisector L1 of the AB side In the triangle ABC, the vertical bisector L1 of AB side intersects BC at D, L2 of AC side intersects BC at e, Li and L2 intersect at point O. the perimeter of triangle ade is 6 and that of OBC of triangle is 16. Find the length of Ao

As shown in the figure, in the triangle ABC, the vertical bisector L1 of the AB side In the triangle ABC, the vertical bisector L1 of AB side intersects BC at D, L2 of AC side intersects BC at e, Li and L2 intersect at point O. the perimeter of triangle ade is 6 and that of OBC of triangle is 16. Find the length of Ao

? the vertical bisector L1 of the AB side intersects with the vertical bisector L2 of the AC side, and the intersection BC is at e ? AE = CE  AD + de + EA = BD + de + EC = BC ? ? adede perimeter is 6 ? BC = 6 ? the vertical bisector L1 of AB edge and L2 of AC edge intersect at the point O  OA = ob = OC ? ob + OC + BC = 2oa + BC = 2oa + 6

As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 120 ° and EF is the vertical bisector of ab. EF intersects BC at point F and ab at point E. verification: BF = 1 2FC．

Proof: connect AF,
∵AB=AC，∠BAC=120°，
∴∠B=∠C=30°，
∵ EF is the vertical bisector of ab,
∴BF=AF，
∴∠BAF=∠B=30°，
∴∠FAC=120°-30°=90°，
∵∠C=30°，
∴AF=1
2CF，
∵BF=AF，
∴BF=1
2FC．

As shown in the figure, △ ABC, ab = AC, ∠ BAC = 120 ° and the vertical bisector EF of AC intersects AC at point E and BC at point F. verification: BF = 2cf

It is proved that: connect AF, (1 minute) ∵ AB = AC,  BAC = 120 °,  B = ∵ C = 180 ° − 120 ° 2 = 30 °, (1 minute)

As shown in the figure, △ ABC, ab = AC, ∠ BAC = 120 ° and the vertical bisector EF of AC intersects AC at point E and BC at point F. verification: BF = 2cf

It is proved that: connect AF, (1 minute) ∵ AB = AC,  BAC = 120 °,  B = ∵ C = 180 ° − 120 ° 2 = 30 °, (1 minute)

As shown in the figure, in △ ABC, ad bisects ∠ BAC, and the vertical bisector EF of ad intersects with the extension line of BC at point F and connects AF. it is proved that ∠ CAF = ∠ B

It is proved that: ∵ EF vertical bisection ad,  AF = DF,  ADF = ∠ DAF,
∵∠ADF=∠B+∠BAD，
∠DAF=∠CAF+∠CAD，
And ∵ ad bisection ∵ BAC,
∴∠BAD=∠CAD，
∴∠CAF=∠B．

As shown in the figure, in △ ABC, ad bisects ∠ BAC, the vertical bisector of ad intersects ad at e, and the extension line intersects BC at F, ∠ B = 40 ° to find the degree of ∠ caf

∠CAF=40°
∠adf=∠b+∠bad
∠ ADF = ∠ DAF (∵ vertical bisector)
∠daf=∠dac+∠caf
∵∠bad=∠dac
∴∠caf=∠b=40°

As shown in the figure, ad is the angular bisector of △ ABC, and the vertical bisector of ad intersects the extension line of BC at point F Verification: fac = ∠ B

It is proved that ∵ EF is the vertical bisector of AD,
∴AF=DF，
∴∠FAD=∠FDA，
∵∠FAD=∠FAC+∠CAD，∠FDA=∠B+∠BAD，
∵ ad bisection ∵ BAC,
∴∠BAD=∠CAD，
∴∠FAC=∠B．

As shown in the figure, point C is on ⊙ o with ab as diameter, CD ⊥ AB is at point P, and AP = a, Pb = B (1) Find the length of string CD; (2) If a + B = 10, find the maximum value of AB, and find the value of a, B at this time

(1) ∵ AB is the diameter of ⊙ o, CD ⊥ AB is at point P,
In the right triangle ACB, according to the projective theorem, PC2 = AP · Pb,
∵AP=a，PB=b，
∴CD=2PC=2
PC2=2
ab，
（2）∵a+b=10，
∴ab≤(a+b
2) 2 = 25, if and only if "a = b = 5"

Point C is on a circle O with diameter AB, CD is perpendicular to P, and let AP = a, Pb = B (1) Find the length of chord CD (2) if a + B = 10, find the maximum value of AB and find the value of a and B at this time Everybody help quickly! Thank you

One condition is missing
If CD ⊥ AB is in P, it will be complete. I will do it according to this
One
CP / AP = BP / CP. CP = √ ab
So CD = 2 √ ab
Two
02√ab≤10
ab≤25
AB max = 25
(10-b)b=25
B=5
A=5
So when a = b = 5, AB reaches the maximum value of 25

In the square ABCD, ∠ man = 45 ° and ∠ man rotates clockwise around point A. its two sides intersect CB, DC at point m, N, connect BD, intersect am with E, and intersect an with F It has been proved that EF squared = be squared + DF squared. Another is that the area of triangle amn is twice that of triangle AEF

Connect ne, make a vertical line from point F to am, and connect MF with vertical foot G
∵∠MAN=∠BDC=45°
∠ AFE = ∠ DFN (common angle)
∴△AEF∽△DFN
∴AF：DF=EF：FN
In △ AFD and △ EFN
∵∠EFN=∠AFD
∴△EFN∽△AFD
∴∠FEN=∠DAN
∵∠AEF=∠DNA
And ∠ Dan + ∠ DNA = 90 degrees
﹤ AEF + ∠ Fen = 90 °
∴NE⊥AM
∵∠MAN=45°
AEN is isosceles right triangle
∴AE=EN
In the same way, we can prove that abmf has four points in the same circle
∴MF⊥AN
∵∠MAN=45°
The △ MAF is an isosceles right triangle
∴FG=（1/2）AM
S△AEF=(1/2)GF×AE=（1/2）EN×（1/2）AM=（1/4）EN×AM
S△AMN=（1/2）AM×EN
∴S△AEF：S△AMN=（1/4）EN×AM/[（1/2）EN×AM]
∴S△AEF：S△AMN=1/2
(finally, the area is simpler than the product of half of the product of both sides and the sine of the included angle, but I can't pass in the picture at a low level) give the score!