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In △ ABC, the vertical bisectors of AB and AC intersect BC at points E and f respectively. If ∠ BAC = 115 °, then ∠ EAF=______ Degree
The vertical bisectors of AB and AC intersect BC at points E and f respectively, so: (1) ea = EB, then ∠ B = ∠ EAG, let ∠ B = ∠ EAG = x degree, (2) FA = FC, then ∠ C = ∠ FAH, let ∠ C = ∠ FAH = y, because ∠ BAC = 115 ° so x + y + ∠ EAF = 115 ° according to the triangle inner angle sum theorem, x + y + X + y + ∠ EAF = 18
In the triangle ABC, the angle BAC is 130 degrees. The vertical bisector of AB intersects BC at point E, and the vertical bisector of AC intersects BC at point g. calculate the degree of angle gae
emergency
The sum of the internal angles of a triangle is 180 degrees
The angle BAC is 130 degrees
Then ABC + BCA = 50 degrees
Then according to the vertical bisector theorem, then angle B = angle BAE
Angle c = angle CAG
So BAE + CAG = 50 degrees
So angle EAG = BAC (130)_ 50=80
In △ ABC, the vertical bisectors of ∠ C = 90 ° and ab intersect BC at d respectively, and ∠ bad - ∠ DAC = 22.5 ° to find the degree of ∠ B
Because De is the vertical bisector of AB, so BD = ad, so, bad = B
Because ∠ bad - ∠ DAC = 22.5 °, therefore, ∠ DAC = ∠ bad-22.5 ° = ∠ b-22.5 °
As ∠ C = 90 °, so ∠ B + ∠ BAC = 90 °,
That is, ∠ B + ∠ bad + ∠ DAC = ∠ B + ∠ B + ∠ B -- 22.5 ° = 90 °,
So, B = 37.5 degree
As shown in the figure, △ ABC, D is the midpoint of BC, e and F are two points on the edge of AB and AC respectively, ed ⊥ FD, which proves that be + CF > EF
It is proved that: extend FD to point m, make MD = FD, connect BM, EM, ∵ D as the midpoint of BC, ∵ BD = CD, in △ FDC and △ MDB, FD = DM, ≌ FDC ≌ mdbcd = BD, ≌ FDC ≌ MDB (SAS), ≌ BM = CF, and ∵ FD = DM, ed ⊥ MF, ≁ ED is the vertical line of MF ≁ EF = em, in △ EBM, be + BM ＞
As shown in Figure 1, in isosceles △ ABC, ab = AC = a, P is any point on the bottom edge BC, PE ∥ AC intersects AB at e through P, PF ∥ AB intersects AC at F,
(1) Verification: PE + pf = a; (2) if the above isosceles △ ABC is changed to isosceles trapezoid ABCD (as shown in Figure 2), where ad ‖ BC, ab = CD, AC and BD intersect at point O, P is any point on the edge of BC, PF ‖ BD intersects DC at F, PE ‖ AC intersects AB at e, and the diagonal length of trapezoid is a, then the conclusion in (1) is still valid, and the reasons are given
(1) It is proved that: ∵ PE ∥ AC, PF ∥ AB, ∥ EBP = ∥ C, quadrilateral aepf is parallelogram, ∥ pf = AE, known isosceles △ ABC, ∥ EPB = ∥ C = ∥ B, ∥ PE = be, ∥ PE + pf = be + AE = AB, ∥ PE + PF = A. (2) (1) is also true. Passing through point P, let PG ∥ CD intersect BD at point G, known isosceles trapezoid ABCD, ad ∥ BC, ab = CD, ∥ ABC = ≌ DCB, BC = BC, ≌ ABC ≌ DCB Therefore, the conclusion in (1) is still valid
It is known that: in △ ABC, ab = AC, P is a point on the bottom edge BC, PE ⊥ AB is on e, PF ⊥ AC is on F, CD is the height on the edge AB, and the proof is: CD = PE + PF (proved by the idea of linear equation)
In the triangle BPE, the angle B = 60 degrees, so the angle BPE = 30 degrees, so be = 1 / 2bp, let be = x, then BP = 2x. Similarly, in the triangle cpd, the angle c = 60 degrees, so the angle cpd = 30 degrees, so CD = 1 / 2cp, let CD = y, then CP = 2Y, so the perimeter of the quadrilateral ebcd is 3x + 3Y + de, because the three sides of the triangle ABC are equal, so BC
As shown in the figure, in △ ABC, ∠ C = 90 °, ad bisects ∠ BAC, de ⊥ ab. if de = 5cm, ∠ CAD = 32 °, calculate the length of CD and the degree of ∠ B
∵ ad bisection ∠ BAC, de ⊥ AB, DC ⊥ AC, ∵ CD = de = 5cm, and ∵ ad bisection ∠ BAC, ∵ BAC = 2 ∠ CAD = 2 × 32 ° = 64 °, ∵ B = 90 ° - ∠ BAC = 90 ° - 64 ° = 26 °
In the known △ ABC and △ def, ab = 2cm, BC = 3cm, CA = 4cm, de = 7.5cm, EF = 10cm, FD = 5cm
Are the two triangles similar? Why?
Certification:
∵AB=2,BC=3,CA=4,DE=7.5,EF=10,FD=5
∴AB/DF＝2/5,AC/DE＝3/7.5＝2/5,BC/EF＝4/10＝2/5
∴AB/DF＝AC/DE＝BC/EF
∴△ABD∽△DFE
Given the congruence of triangle ABC and triangle def, and de = 3cm, ab = 4cm, CA = 6cm, find the perimeter of triangle def
The corresponding sides of congruent triangles are equal, and the perimeter is also equal. In △ ABC, we know that the length of two sides AB = 4cm, CA = 6cm, and we also know that de = 3cm in △ def, then this de must be the corresponding side with BC. The perimeter of triangle def is the perimeter of triangle ABC, which is equal to 3 + 4 + 6 = 13cm
Given △ ABC ∽ def, Qie de = 3, ab = 4, BC = 5, CA = 6, find the perimeter of △ def
ABC circumference = 4 + 5 + 6 = 15
Δ ABC perimeter: △ def perimeter = AB: de
15: Δ def perimeter = 4:3
Δ def perimeter = 45 / 4 = 11.25
RELATED INFORMATIONS
- 1. In △ ABC, ∠ C = 90 ° and the intersection line BC of AB is at D. if ∠ bad - ∠ DAC = 22.5 °, then the degree of ∠ B is______ ．
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