As shown in the figure, in diamond ABCD, e is the midpoint of AD, and the extension line of EF ⊥ AC crossing CB is at F Verification: AB and EF are equally divided

As shown in the figure, in diamond ABCD, e is the midpoint of AD, and the extension line of EF ⊥ AC crossing CB is at F Verification: AB and EF are equally divided

Proof: connect BD, AF, be,
In diamond ABCD, AC ⊥ BD
∵EF⊥AC，
/ / EF ∥ BD, ed ∥ FB,
The quadrilateral EDBF is a parallelogram, de = BF,
∵ e is the midpoint of AD,
∴AE=ED，∴AE=BF，
AE ‖ BF,
The quadrilateral aebf is a parallelogram,
That is, AB and EF are equally divided

As shown in the figure, in diamond ABCD, e is the midpoint of AD, and the extension line of EF ⊥ AC crossing CB is at F Verification: AB and EF are equally divided

Proof: connect BD, AF, be,
In diamond ABCD, AC ⊥ BD
∵EF⊥AC，
/ / EF ∥ BD, ed ∥ FB,
The quadrilateral EDBF is a parallelogram, de = BF,
∵ e is the midpoint of AD,
∴AE=ED，∴AE=BF，
AE ‖ BF,
The quadrilateral aebf is a parallelogram,
That is, AB and EF are equally divided

As shown in the figure, in diamond ABCD, e is the midpoint of AD, and the extension line of EF ⊥ AC crossing CB is at point F. it is proved that AB and EF are equally divided

Proof: connect BD, AF, be,
In diamond ABCD, AC ⊥ BD
∵EF⊥AC,
/ / EF ∥ BD, ed ∥ FB,
The quadrilateral EDBF is a parallelogram, de = BF,
∵ e is the midpoint of AD,
∴AE=ED,∴AE=BF,
AE ‖ BF,
The quadrilateral aebf is a parallelogram,
That is, AB and EF are equally divided

In ▽ ABC, BD and CE are two high lines, f is a point on BD, G is a point on CE extension line, BF = AC, CG = ab. please judge the shape of triangle AFG

an isosceles triangle
No pictures! Do you understand?
Because angle a is a common angle, angle AEC = angle ADB = 90 degrees,
So the angle abd = the angle ace
In ▽ ABF and ▽ GCA
BF＝AC
CG＝AB
Angle abd = angle ace
So ▽ ABF and ▽ GCA are congruent
So AF = AG
So ▽ AGF is an isosceles triangle

As shown in the figure, BD, CE are the height of the triangle ABC, the point F is on BD, BF = AC, the point G is on the extension line of CE, CG = AB, try to explain the relationship between Ag and AF, and explain the reasons

The ∵ BD, CE are all the heights of ∵ ABC, ? ACG + ∵ CE are all the height of ? ABC, ? ACG + ∵ BAC = 90 ᙽ ACG ? AC 8780\87808780878087808780878087808780≌ FBA, ? FBA, ? g, ? g = ≌ FBA, ? FBA, ﹤ G + ﹤ gae = 90 °, ﹤ EAF +

Known: as shown in the figure, △ ABC, CE ⊥ AB, BF ⊥ AC

It is proved that ∵ CE ⊥ AB is in E, BF ⊥ AC is in F,
∴∠AFB=∠AEC．
A is the common angle,
∨ Abf ∽ △ ace
ν AB: AC = AF: AE, ∠ A is the common angle
△aef ∽ △ ACB (two triangles with equal angles corresponding to two sides are similar)

As shown in the figure, BD, CE are the height on the edge AC and ab of the triangle ABC, BF = AC, CG = AB, find the relationship between Ag and AF

AG = AF and Ag ⊥ AF
The reasons are as follows: ① AF = AG,
∵ BD and CE are higher than ∵ ABC,
∴∠ACG+∠BAC=90°,∠FBA+∠BAC=90°,
∴∠ACG=∠FBA,
∵BF=AC,CG=AB,
∴△ACG≌△FBA,
∴AF=AG．
②AF⊥AG,
∵△ACG≌△FBA,
∴∠G=∠EAF,
∵CG⊥AB,
∴∠G+∠GAE=90°,
∴∠EAF+∠GAE=90°,
∴AG⊥AF,
⊥ Ag = AF and Ag ⊥ AF
Hope to adopt, if you do not understand, please ask

In ⊙ o with radius 1, the lengths of chord AB and AC are root 2 and root 3 respectively. Find the degree of ⊙ BAC

Because radius = 1, so OA = ob = OC = 1, and because AC = √ 2, so OA ^ 2 + OC ^ 2 = AC ^ 2, so the triangle OAC is an isosceles right triangle, so ∠ Cao = 45 through the center of the circle O as OD ⊥ AB, then D is the midpoint of AB, so ad = √ 3 / 2, so cos ∠ OAB = ad / AO = √ 3 / 2, so ∠ OAB = 30 degrees, so ∠ BAC = 30

Given that the radius of circle O is 1, the length of chord AB and AC is root 2 and root 3, then the degree of angle BAC is?

Connected with OA and ob
OA=OB=1
So, OA: OB: ab = 1:1: radical 2
so,∠OAB=45°
Make OD ⊥ to AC
So, ad = root of half 3
Because OA = 1
So ∠ oad is equal to 30 degrees
so,∠CAB=45°+30°=75°

In ⊙ o with radius 5, the chord AB = 5 2, the chord AC = 5, then the degree of ∠ BAC is___ .

As shown in the figure, OC, OA, ob.  OC = OA = AC = 5,  OAC is an equilateral triangle,  Cao = 60 °, ∵ OA = ob = 5, ab = 52,  oa2 + ob2 = 50 = AB2,  OAB is an isosceles right triangle,