It is known that the bisection angle BAC, De is perpendicular to AB and E, DF is perpendicular to AC and F, DB = DC

It is known that the bisection angle BAC, De is perpendicular to AB and E, DF is perpendicular to AC and F, DB = DC

prove:
∵ ad is the bisector of ∵ BAC,
De is perpendicular to AB and DF is perpendicular to AC
∴ DE=DF
And ∵ DB = DC
The right triangle △ bed is similar to △ CFD
∴∠B=∠C
The ABC is an isosceles triangle

As shown in the figure, ∠ C = 90 ° in △ ABC (1) Find a point D on BC so that the distance from point d to AB is equal to the length of DC; (2) Connect ad, draw a triangle and △ ABC symmetric about the line ad

(1) As shown in the figure: point D is the required value;
(2) As shown in the figure: △ AFE and △ ABC are symmetric about the straight line ad

As shown in the figure, in △ ABC, ∠ BAC = 120 °, ad ⊥ BC is in D, and ab + BD = DC, then ∠ C=______ Degree

Intercept de = dB on DC and connect AE,
Let ∠ C = X,
∵AB+BD=DC，DE=DB，
∴CE=AB，
And ∵ ad ⊥ BC, DB = De,
The straight line ad is the vertical bisector of be,
∴AB=AE，
∴CE=AE，
∴∠B=∠AEB，∠C=∠CAE，
And ∵ AEB = ∠ C + ∠ CAE,
∴∠AEB=2x，
∴∠B+∠C=3x=180°-120°=60°，
∴∠C=20°．
So the answer is: 20 degrees

As shown in the figure, in △ ABC, ∠ BAC = 120 °, ad ⊥ BC is in D, and ab + BD = DC, then ∠ C=______ Degree

Intercept de = dB on DC and connect AE,
Let ∠ C = X,
∵AB+BD=DC，DE=DB，
∴CE=AB，
And ∵ ad ⊥ BC, DB = De,
The straight line ad is the vertical bisector of be,
∴AB=AE，
∴CE=AE，
∴∠B=∠AEB，∠C=∠CAE，
And ∵ AEB = ∠ C + ∠ CAE,
∴∠AEB=2x，
∴∠B+∠C=3x=180°-120°=60°，
∴∠C=20°．
So the answer is: 20 degrees

In △ ABC, ab = AC, ad ⊥ BC, points E and F are the midpoint of BD and DC respectively, then the congruent triangles in the graph share () A. 3 pairs B. 4 pairs C. 5 pairs D. 6 pairs

∵AD⊥BC，AB=AC，
ν D is the midpoint of BC,
∴BD=DC，
∴△ABD≌△ACD（HL）；
∵ E and F are the midpoint of DB and DC respectively,
∴BE=ED=DF=FC，
∵AD⊥BC，AD=AD，ED=DF，
∴△ADF≌△ADE（HL）；
∵∠B=∠C，BE=FC，AB=AC，
∴△ABE≌△ACF（SAS），
∵EC=BF，AB=AC，AE=AF，
∴△ABF≌△ACE（SSS）．
≌△ ACD (HL), △ Abe ≌ △ ACF (SAS), △ ADF ≌ △ ade (SSS), △ Abf ≌ △ ACE (SAS)
Therefore, B

In the triangle ABC, Bo bisection angle ABC, CO bisection angle ACB, de through O and parallel BC, triangle ade perimeter = 10cm, BC = 5cm, calculate the circumference of triangle ABC It's urgent! Must have the concrete step!!! T－T

So the triangle dbo is an isosceles triangle BD = do. Similarly, EO = EC △ ade perimeter = AD + AE + do + EO = 10 △ ABC perimeter = AD + AE + BD + CE + BC = AD + AE + do + EO + BC = AD + AE + do + EO + BC = AD + AE + do + EO + BC = 10 + 5 = 15cm

As shown in the figure, in △ ABC, Bo bisects ∠ ABC, CO bisects ∠ ACB, de passes through O and is parallel to BC. It is known that the circumference of △ ade is 10cm and the length of BC is 5cm. Calculate the circumference of △ ABC

∵ Bo bisection ∠ ABC, CO bisection ∠ ACB,
∴∠DBO=∠OBC，∠ECO=∠OCB，
∵DE∥BC，
∴∠DOB=∠OBC，∠EOC=∠OCB，
∴∠DBO=∠DOB，∠ECO=∠EOC，
ν BD = OD, CE = EO (equiangular to equilateral)
∵AD+DE+AE=10cm，
∴AD+BD+CE+EA=10cm，
And BC is 5cm long, so the circumference of △ ABC is:
AD+BD+CE+EA+BC=10+5=15cm．

As shown in the figure △ ABC, Bo bisects ∠ ABC, CO bisects ∠ ACB, de passes through O and is parallel to BC. It is known that the circumference of △ ade is 12cm and BC is 5cm. Calculate the circumference of △ ABC

∵ Bo bisection ∠ ABC, CO bisection ∠ ACB,
∴∠ABO=∠OBC，∠ACO=∠OCB，
∵DE∥BC，
∴∠BOD=∠OBC，∠COE=∠OCB，
∴∠ABO=∠BOD，∠ACO=∠COE，
∴BD=OD，CE=OE，
The circumference of ade is 12cm,
∴AD+DE+AE=AD+OD+OE+AE=AD+BD+CE+AE=AB+AC=12cm，
∵ BC length is 5cm,
∴AB+AC+BC=17（cm），
The circumference of △ ABC is 17 cm

In the triangle ABC, D is the midpoint of AC, BD is perpendicular to AC, De is parallel to BC, AB intersects with point E, BC = 5cm, AC = 4cm, calculate the perimeter of triangle ade

7cm
D is the midpoint of AC, BD is perpendicular to AC,
Ba = BC = 5
D is the midpoint of AC and De is parallel to BC
So De is the median line and E is the midpoint of ab
Therefore, the perimeter of the triangle ade is equal to AE + ed + da
=AB/2+BC/2+CA/2
=(5+5+4)/2
=7cm

As shown in the figure, Ao and Bo bisect ∠ cab and ∠ CBA respectively, and the distance from point O to ab od = 2cm, and the circumference of △ ABC is 14cm, then △ ABC area is equal to

O is OE ⊥ BC, of ⊥ AC;
∵ AO and Bo are equally divided ᙽ cab, ᙽ CBA respectively;
∴OD=OE=OF；
The area of △ ABC = 1 / 2 × OD × AB + 1 / 2 × OE × BC + 1 / 2 × of × AC
=1/2×OD×(AB+BC+AC)
=1/2×2×14
=14cm^2