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In the triangle ABC, the bisector of angle B and the bisector of angle c intersect at point DDG / BC, and acab is proved to be GF = bg-cf at two points of FG
Because dg / / BC
So ∠ GDB = ∠ DBC
Because BD is an angular divider
Therefore, abd = DBC, so GDB = GBD, BG = GD
Because CD is an angular divider dg / / BC
So ∠ GDC = ∠ FCD
So CF = DF
Because GF = gd-fd
So GF = bg-cf
As shown in the figure, △ ABC, ab = AC, points D and E are on the extension line of AB and AC respectively, and BD = CE, de and BC intersect at point F. verification: DF = EF
It is proved that DG ∥ AE is made through point D and passed to point BC at point G, as shown in the figure,
∴∠1=∠2，∠4=∠3，
∵AB=AC，
∴∠B=∠2，
∴∠B=∠1，
∴DB=DG，
And BD = CE,
∴DG=CE，
In △ DFG and △ EFC
∠4＝∠3
∠DFG＝∠EFC
DG＝CE ，
∴△DFG≌△EFC，
∴DF=EF．
As shown in the figure, in △ ABC, given that ab = AC, take point D on AB, and take point E on the extension line of AC, so that CE = BD, connect de and cross BC to g, then DG = Ge, why In △ ABC, it is known that ∠ ABC = 60 °, ACB = 45 °, ad, CF are the heights on the sides of BC and AB, and intersect at point P, the angular bisector be of ∠ ABC intersects ad, CF at m, n respectively. Try to find out all isosceles triangles in the figure, and briefly describe the reasons There are two questions in total. They are all from the Experimental Manual of the first semester of junior high school on page p24. Choose the answer in 30 minutes,
1. Make DF parallel BC to AC to F,
Angle ABC = angle ACB,
Angle ADF = angle AFD,
AD=AF,
BD=CF,
CF=CE,
GC is parallel to DF, GC is the median line of triangle EDF,
DG=GE.
2. Triangle ADC is isosceles right triangle. Angle ACB = angle DAC = 45 degrees
The triangle PMN is an equilateral triangle. Angle ABN = 30 degrees, angle fnb60 degrees, angle PMN = angle BMD = 90 degrees, angle CBN = 60 degrees
I wish you progress in your study
It is known that in the triangle ABC, take a point D on AB, and take a point E on the AC extension line, so that CE = BD, connect De to BC, and have DG = Ge
Make ef parallel BD intersection BC extension line at F
Because DG = Ge, and BD is parallel to ef
Easy to know △ BDG congruence △ EFG
There is BD = EF
It is also known that CE = BD
So EC = EF
Because AB is parallel to ef
So △ ABC is similar to △ CEF
There are ab: AC = EF: EC = 1
So AB = AC
As shown in the figure, △ ABC, ab = AC, points D and E are on the extension line of AB and AC respectively, and BD = CE, de and BC intersect at point F. verification: DF = EF
It is proved that DG ∥ AE is made through point D and passed to point BC at point G, as shown in the figure,
∴∠1=∠2，∠4=∠3，
∵AB=AC，
∴∠B=∠2，
∴∠B=∠1，
∴DB=DG，
And BD = CE,
∴DG=CE，
In △ DFG and △ EFC
∠4＝∠3
∠DFG＝∠EFC
DG＝CE ，
∴△DFG≌△EFC，
∴DF=EF．
As shown in the figure, △ ABC, ab = AC, points D and E are on the extension line of AB and AC respectively, and BD = CE, de and BC intersect at point F. verification: DF = EF
0
0
It is proved that DG ∥ AE is made through point D and passed to point BC at point G, as shown in the figure,
∴∠1=∠2，∠4=∠3，
∵AB=AC，
∴∠B=∠2，
∴∠B=∠1，
∴DB=DG，
And BD = CE,
∴DG=CE，
In △ DFG and △ EFC
∠4＝∠3
∠DFG＝∠EFC
DG＝CE ，
∴△DFG≌△EFC，
∴DF=EF．
As shown in the figure, △ ABC, ab = AC, points D and E are on the extension line of AB and AC respectively, and BD = CE, de and BC intersect at point F. verification: DF = EF
0
As shown in the figure, it is known that AB is the chord of ⊙ o, OB = 4, ∠ OBC = 30 ° and point C is any point on chord AB (not coincident with points a and b), connect CO and extend Co, intersect ⊙ o with point D, and connect AD and DB (1) When ∠ ADC = 18 °, calculate the degree of ∠ DOB; (2) If AC = 2 3. Verification: △ ACD △ OCB
(1) Connect OA,
∵OA=OB=OD，
∴∠OAB=∠OBC=30°，∠OAD=∠ADC=18°，
∴∠DAB=∠DAO+∠BAO=48°，
According to the circular angle theorem, it is concluded that: ∠ DOB = 2 ∠ DAB = 96 °
(2) It is proved that o e ⊥ AB is at point E, and the perpendicular foot is e,
∵ OE over O,
According to the vertical diameter theorem, AE = be,
∵ in RT △ OEB, OB = 4, ∵ OBC = 30 °,
∴OE=1
2OB=2，
From be = 2, the Pythagorean theorem is obtained
3=AE，
AB = 2ae = 4
3，
∵AC=2
3，
∴BC=2
3，
That is, the two points of C and e coincide,
∴DC⊥AB，
∴∠DCA=∠OCB=90°，
∵DC=OD+OC=2+4=6，OC=2，AC=BC=2
3，
∴AC
OC=CD
BC=
3，
The two triangles with the same angle are similar
As shown in the figure, the relationship between the mass and volume of two substances a and B is shown. If solid cylinders a and B with equal mass are made into equal height solid cylinders a and B, and they are placed side by side on the horizontal ground, the pressure ratio of two cylinders a and B to the horizontal ground is () A. 8：1 B. 4：3 C. 4：1 D. 1：2
(1) It can be seen from the figure that when v a = 1cm3, m a = 8g; when m b = 4G, v b = 4cm3, then ρ a = m a, V A = 8g / cm3, ρ B = M B, v b = 4g4cm3 = 1g / cm3, ρ a ρ B = 8g / cm3, ρ a ρ B = 8g / cm3 = 81, (2) ∵ the pressure of the cylinder on the horizontal ground P = FS = GS = MGS = ρ VGS = ρ sh
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