It is proved that ab minus AC is greater than BD minus DC

It is proved that ab minus AC is greater than BD minus DC

∵ ab ∵ AC, take a point E on AB so that AE = AC, connected with de. ∵ ad is the bisector of ∵ BAC,  DAE = ≌ △ ADC, (s, a, s), ≌ de = CD, in the triangle bed, be = ab-ae = AB-AC, bd-dc = bd-de, there are be ﹥ bd-de, ≌≌△ ADC, (s, a, s), ∵ de = CD, in the triangle bed, be = ab-ae = AB-AC, bd-dc = bd-de, there are be ﹥ bd-de, ﹥ AB-AC ≌

It is known that in the triangle ABC, ad is the bisector of ∠ A. It is proved that ab: AC = BD: DC

It is proved that the extension line of crossing point C is CE ∥ AB and the extension line of ad is at E
∵ ad bisection ∵ bac
∴∠BAD＝∠CAD
∵CE∥AB
∴∠E＝∠BAD
∴∠E＝∠CAD
∴CE＝AC
And ∵ CE ∵ ab
∴△ABD∽△ECD
∴AB/CE＝BD/CD
∴AB/AC＝BD/CD
∴AB:AC=BD:DC

Ad is the angular bisector of the triangle ABC, and BD = DC To pay,

Because ad is the angular bisector of the triangle ABC, and BD = DC, so ad "three lines in one"
So the triangle ABC is an isosceles triangle, so AB = AC

It is known that in the triangle ABC, ad is a bisector of ∠ A. It is proved that ab: AC = BD: DC

D is de vertical AB, D is DF vertical AC,
Through a, Ag is perpendicular to BC,
Because ad is a bisector, there is de = DF
The area of triangle abd = 1 / 2 * AB * de = 1 / 2 * BD * AG
AB / BD = Ag / De
Area of triangle ADC = 1 / 2 * ac * DF = 1 / 2 * DC * AG
AC / DC = Ag / DF
So AB / BD = Ag / de = Ag / DF = AC / DC (because de = DF)
So AB / BD = AC / DC, that is, AB: AC = BD: DC

As shown in Figure 12, △ ABC is an equilateral triangle, D and E are points on BC and AC respectively, and BD = CE, ad and be intersect at point P, and the degree of ∠ ape is calculated

Firstly, it is proved that △ abd and △ BCE are congruent, and ∠ bad = ∠ CBE is obtained
∠ ape = ∠ Abe + ∠ bad = ∠ APE + ∠ CBE = 60 degrees

As shown in the figure, △ ABC is an equilateral triangle, CE is an exterior bisector, point D is on AC, connecting BD and extending its intersection with CE at point E (1) Verification: △ abd ∽ CED (2) If AB = 6, ad = 2CD, find sin ∠ EBC

(1) It is proved that ∵ △ ABC is an equilateral triangle,
∴∠A=∠ACB=60°，
∵ CE is the bisector of ∵ ACF
∴∠ACE=∠A=60°，
ADB = ∠
∴△ABD∽△CED；
(2) Make DH ⊥ BC at point H,
∵∠ACB=60°，
∴∠HDC=30°
∵AC=6，AD=2CD，
∴CD=2，AD=4，
∵∠HDC=30°，
∴HC=1
2DC=1，DH=
3，BH=6-1=5，
∴BD=
25+3=2
7，
∴sin∠EBC=DH
BD=
Three
Two
7=
Twenty-one
14．

As shown in Fig. 1, in the triangle ABC, ab = AC, point D is the midpoint of BC, and point E is on ad. (1) verification: be = CE. (2) as shown in Fig. 2, if the extension line of be intersects AC at point F In addition, BF is perpendicular to AC, foot is f, angle BAC is 45 ° and other conditions remain unchanged. It is proved that AEF of triangle is equal to BCF of triangle

(1)
∵ AB = AC, D is the midpoint of BC
∴AD⊥BC,BD=CD
∴△BDE≌△CDE
∴BE=CE
（2）
∵BF⊥AC,
∴∠BFC=∠AEF=90°
And ∵ BAC = 45 
△ ABF is an isosceles triangle AF = BF
∵∠C+∠CBF=90°
∠C+∠EAF=90°
∴∠CBF=∠EAF
∴△AEF≌△BCF（ASA)

Let a be a vertex of a graph, such as a C, a B C, and a B C, respectively

Solution; △ ABC is an equilateral triangle,
∴AB=BC，∠ABC=∠C=60°．
In △ abd and △ BCE,
AB＝BC
∠ABD＝∠BCE
BD＝CE ，
∴△ABD≌△BCE（SAS），
∴∠BAD=∠CBE．
From the properties of the triangle angle, we can get ∠ AFE = ∠ BAF + ∠ ABF,
∠AFE=∠CBE+∠ABF=60°．

Let a be a vertex of a graph, such as a C, a B C, and a B C, respectively

Solution; △ ABC is an equilateral triangle,
∴AB=BC，∠ABC=∠C=60°．
In △ abd and △ BCE,
AB＝BC
∠ABD＝∠BCE
BD＝CE ，
∴△ABD≌△BCE（SAS），
∴∠BAD=∠CBE．
From the properties of the triangle angle, we can get ∠ AFE = ∠ BAF + ∠ ABF,
∠AFE=∠CBE+∠ABF=60°．

Is the vertex of a triangle, which is equal to the degree of △ E

In triangle ACD and triangle CBE, AC = CB; CD = be; ㄥㄥ C = ㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥㄥ