In the triangle ABC, ∠ BAC is the acute angle, h is the intersection of high AD and be, ad = BD, BH = AC

In the triangle ABC, ∠ BAC is the acute angle, h is the intersection of high AD and be, ad = BD, BH = AC

In RT △ AEH and RT △ ADC
∠AHE=90-∠DAC
∠C=90-∠DAC
Therefore, ahe = C
Because ∠ ahe = ∠ Bhd (opposite vertex angle)
So ∠ C = ∠ Bhd (equivalent substitution)
In RT △ BDH and RT △ ADC
∠HDB=∠ADC=90
Ad = BD (known)
∠ C = ∠ Bhd (proven)
So RT △ BDH and RT △ ADC are identical
So BH = AC

As shown in the figure, in the triangle ABC, the angle ABC = 45 degrees, ad is perpendicular to BC at point D, be is perpendicular to AC at point E, ad intersects be at point H

Proof: in the isosceles right triangle abd, ad = BD; right triangle ADC and right triangle AEH; angle hea = angle CDA, angle hae = angle CAE, (equal to the vertex angle) triangle hea is similar to triangle CDA; so angle H = angle c; because: angle H = angle c, angle hea = angle CDA, ad = BD, so: Triangle HBD is equal to triangle CAD; (angle side) BH = AC; proof

As shown in the figure, in △ ABC, ad ⊥ BC is at D, be ⊥ AC is at e, and AD and be intersect at point F. if BF = AC, what is the size of ⊥ ABC?
The process to be deduced is easy to understand

∵AD⊥BC,BE⊥AC,
∴∠BDF=∠AEB=90°
The mutual complements of ∠ DBF and ∠ DFB, EAF and AFE,
And ∠ DFB = ∠ AFE
∴∠DBF=∠EAF
And ∵ BF = AC
∴ΔBDF≌ΔCAD
∴BD=AD
That is, Δ abd is an isosceles right triangle,
Therefore, ABC = 45 degree

As shown in the figure, if three circles with radius 3 are circumscribed, and each side of △ ABC is tangent to two of them, then the perimeter of △ ABC is___ ．

As shown in the figure, ∵ connect Ao, Op, Pb, OE, PF, on; ∵ according to the properties of tangent two circles, Op = PN = on = 23, ∵ ONP is an equilateral triangle, ∵ OPN = ∵ PON = ∵ ONP = 60 °, ∵ according to the properties of tangent, OE ⊥ AB, PF, ∵ OE ∥ PF, OE = PF, ∵ quadrilateral oefp is a rectangle, ∥ op ∥ AB, the same as PN ∥ BC, on ∥ AC, then ∵ OPN = ∵ ABC = 60 ° and ∵ PON = ∵ BAC = 60 ° according to the properties of tangent In RT △ AOE, ∠ EAO = 30 ° and OE = 3; then AE = 3, BF = 3; because ⊙ O and ⊙ P are circumscribed, Op = 23; so AB = AE + EF + BF = 6 + 23; according to tangent length theorem, ab = BC = AC, so the circumference of △ ABC is 18 + 63

What is the circumference of a triangle if three circles of radius 3 are circumscribed and inscribed on the triangle ABC?

Therefore, the triangle ABC is an equilateral triangle, in which the centers of the three circles pass through the angle bisector of the corresponding angle, and each center has two radii perpendicular to the sides of ABC, forming a right angle triangle of 30 °, 60 ° and 90 °, one of which is the radius root 3

If three circles with radius root 3 are tangent to each other and each side of triangle ABC is tangent to two of them, what is the circumference of triangle ABC?

The distance from a to the first tangent point is: √ 3 / Tan 30 ° = 3 √ 3, and the distance from B to the tangent point is also = 3 √ 3. The distance from the remaining part of AB side = 2R = 2 √ 3, so the length of AB side = 3 √ 3 × 2 + 2 √ 3 = 8 √ 3 Perimeter = 8 √ 3 × 3 = 24 √ 3

As shown in the figure, if three circles with radius 3 are circumscribed, and each side of △ ABC is tangent to two of them, then the perimeter of △ ABC is___ ．

As shown in the figure, ∵ connect Ao, Op, Pb, OE, PF, on; ∵ according to the properties of tangent two circles, Op = PN = on = 23, ∵ ONP is an equilateral triangle, ∵ OPN = ∵ PON = ∵ ONP = 60 °, ∵ according to the properties of tangent, OE ⊥ AB, PF, ∵ OE ∥ PF, OE = PF, ∵ quadrilateral oefp is a rectangle, ∥ op ∥ AB, the same as PN ∥ BC, on ∥ AC, then ∵ OPN = ∵ ABC = 60 ° and ∵ PON = ∵ BAC = 60 ° according to the properties of tangent In RT △ AOE, ∠ EAO = 30 ° and OE = 3; then AE = 3, BF = 3; because ⊙ O and ⊙ P are circumscribed, Op = 23; so AB = AE + EF + BF = 6 + 23; according to tangent length theorem, ab = BC = AC, so the circumference of △ ABC is 18 + 63

Given that three circles with radii of root 3 are circumscribed, and each side of triangle ABC is inscribed with two of them, the perimeter of triangle ABC is calculated

∵ three circles are tangent to each other, so the circumscribed △ ABC is an equilateral triangle,
Ψ Bo2 bisection ∠ ABC, ∠ o2bc = 30 °
∵O2D⊥BD
∴O2D/BD＝tan30°＝（√3）/3
∴BD＝O2D/〔（√3）/3〕＝（√3）/〔（√3）/3〕＝3
Similarly, CE = 3
DE＝O2O3＝2√3
∴BC＝BD+DE+EC＝3+3+2√3＝6+2√3
The circumference of ∧ ABC is:
3BC＝3×（6+2√3）＝18+6√3

As shown in the figure, in RT △ ABC, we can find the inscribed circle radius of right triangle by ∠ C = 90 ° AB = 6 and AC = 4

Let the radius of the inscribed circle of a triangle be r, then s △ ABC = 1 / 2 * ac * BC = 1 / 2 * (AB + BC + AC) * r
It can be obtained from ∠ C = 90 ° AB = 6, AC = 4 and Pythagorean theorem in RT △ ABC
BC=2√5
Then s △ ABC = 1 / 2 * ac * BC = 1 / 2 * (AB + BC + AC) * r can be transformed into
r=(AC*BC)/(AB+BC+AC)=√5-1

In the RT triangle ABC, ∠ ACB is equal to 90 ° AC = BC, D is on AC, and ∠ CDB = 60 ° to find the value of ad divided by CD

AC = BC, DC = radical 3 / 3bC, ad ratio CD = (1-radical 3 / 3) / radical 3 = (radical 3-1) / 3