It is known that, as shown in the figure, in the triangle ABC, ab = AC, point D is the midpoint of BC, De is perpendicular to AB and E, DF is perpendicular to AC and F, so it is proved that angle DFE = angle def

It is known that, as shown in the figure, in the triangle ABC, ab = AC, point D is the midpoint of BC, De is perpendicular to AB and E, DF is perpendicular to AC and F, so it is proved that angle DFE = angle def

∵AB=AC
∴∠B=∠C
And ∵ de ⊥ AB, DF ⊥ AC
∴∠BDE=∠CDF
And ∵ BD = DC
∴ΔBDE≌ΔCDF
∴BE=CF,AE=AF
∴EF//BC
∴∠DFE=∠CDF, ∠DEF=∠BDE
∴∠DFE=∠DEF

In the triangle ABC, Ba is equal to Ca and 13 cm. The bisector of Ba intersects AB at D and AC at E. if the circumference of the triangle EBC is 21 cm, then
In the triangle ABC, Ba is equal to Ca, 13 cm. The bisector of Ba intersects AB at D and AC at E. if the circumference of the triangle EBC is 21 cm, what is BC equal to?

BC = 8 cm
Because De is the vertical bisector of △ AEB, AE = EB
Because the circumference of the triangle EBC is 21 cm
So be + EC + BC = 21 cm
Because be = AE, be + EC + BC = AE + EC + BC = 21cm, that is, AC + BC = 21cm
Then BC = 21-ae = 21-13 = 8 cm

In △ ABC, ab = AC, the angle of the vertical bisector of AB intersects AB at D, AC and E. if the perimeter of △ ABC and △ EBC is 26cm.18cm respectively, the length of AC is calculated
The answer is 10, but I can't do it

∵ AB = AC, △ ABC circumference is 26cm
∴2AC+BC=26
∵ e is on the vertical bisector of ab
∴AE=BE
∵△ BCE perimeter = BC + CE + be = BC + CE + AE = BC + AC = 18
∴AC=8cm．

Triangle ABC and D are the vertical bisectors of AB, which intersect AC with E and connect be. The perimeter of triangle EBC is 26cm and AC is 16C
m. Seeking BC length

Because the triangle BDE and ade will wait, be = AE, BC = 26-16 = 10

E is a point in △ ABC. Try to prove that: ∠ aebd = ∠ C ＋ CAE ＋ CBE

Proof: connect CE and extend to F,
From the outside angle of the triangle,
∠AEF＝∠EAC+∠ACE
∠BEF＝∠EBC+∠BCE
Therefore, AEF + bef = EAC + ACE + EBC + BCE
That is: ∠ AEB ＝ C ＋ CAE ＋ CBE

As shown in the figure, in the triangle ABC, angle a = half angle c = half angle ABC
I don't need to play outside. I haven't learned yet

Therefore, a = 36 degrees. ABC = C = 72 degrees,
BDC = 180-72-72 / 2 = 72 degrees

A. P, B and C are the four points on the circle O, ∠ APC = ∠ CPB = 60 ° to judge the shape of triangle ABC

Equilateral triangle! Because the Xuan AC of the angle APC pair is equal to the Xuan BC of the angle CPB pair! And because the angle APB is 120 degrees (AB does not coincide) or 60 degrees (point a and point B coincide)! That is, the angle ACB is 60 degrees! So the triangle ABC is an equilateral triangle

As shown in the figure, the four points of a, P, B and C on the circle O, ∠ APC = ∠ CPB = 60 °, judge the shape of △ ABC and prove your conclusion

A: ABC is an equilateral triangle
It is proved that ∵ CPB and ∵ cab are in the same arc and ∵ CPB = 60 degree
∴∠CAB=∠CPB=60°
And ∵ APC and ∵ ABC are in the same arc, and ∵ APC = 60 degree
∴∠ABC=∠APC=60°
Then, in △ ABC, ABC = cab = 60 degree
The ABC is an equilateral triangle

It is known that as shown in the figure, ∠ APC = ∠ CPB = 60 ° to prove that △ ABC is an equilateral triangle

It is proved that: ∵ AC = AC, ∵ ABC = ∵ APC = 60 °. Similarly, ∵ BAC = ∵ CPB = 60 °, then ∵ ACB = 180 ° - ∵ ABC - ∵ BAC = 60 °, so △ ABC is an equilateral triangle

A. P, B and C are the four points on the circle O, ∠ APC = ∠ CPB = 60 degrees. Judge the shape of △ ABC and prove the conclusion

So AC = BC is connected with OA OC oboa = OC = ob △ AOC is isosceles triangle ∠ AOC = 2 ∠ OCA = 120 °∠ OAC = ∠ OCA = 30 ° the same as ∠ OBC = ∠ OCB = 30 °∠ ACB = ∠ OCA = ∠ OCB = 30 °∠ ACB = ∠ OCA + ∠ OCB = 30 + 30 = 60 °