O is the center of the triangle ABC circumscribed circle, connecting Ao, BC, D, Bo, AC, e, Co, AB and F. R is the radius. The proof is: 1 / AD + 1 / be + 1 / CF = 2 / R

Let ha, Hb and HC be the heights of △ ABC edges BC, Ca and ab respectively, BC = a, CA = B, ab = C, and s denote their area. ∵ 1 / ad = cos (B-C) / ha, 1 / be = cos (C-A) / Hb, 1 / CF = cos (a-b) / HC, s = (Sina * BC) / 2 = 2R ^ 2 * 4sina * SINB * sinc ∵ 1 / AD + 1 / be + 1 / CF = a * cos (B-C) / (2S) + b * cos (C-A) / (2

The outer center of acute triangle △ ABC is O, the radius of circumscribed circle is r, extending Ao, Bo, Co, intersecting with opposite sides BC, CA, AB at D, e, f respectively; it is proved that 1AD + 1Be + 1CF = 2R

It is proved that ad, be and CF are common o, odad = s △ OBCS △ ABC, oebe = s △ oacs △ BAC, OFCF = s △ oabs △ cab 5 ', then odad + oebe + OFCF = 1 ①… And odad = R − dm2r − DM = 1 − R2R − DM = 1 − rad Similarly, oebe = 1 − RBE, & nbsp; OFCF = 1 − RCF Substituting 20 'into (1), we can get (1 − RAD) + (1 − RBE) + (1 − RCF) = 1 ② So & nbsp; 1AD + 1Be + 1CF = 2R. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp 25’

The acute triangle ABC is connected in the circle, and Ao, Bo and Co are connected with BC, AC and AB to D, e and f respectively, and 1 / AD + 1 / be + 1 / CF = 2 / R is proved

Let ha, Hb and HC be the heights of △ ABC, BC, Ca and ab respectively, BC = a, CA = B, ab = C, and s denote their area
∵1/AD=cos(B-C)/ha,1/BE=cos(C-A)/hb,1/CF=cos(A-B)/hc
S=(sinA*bc)/2=2R^2*4sinA*sinB*sinC
∴1/AD+1/BE+1/CF
=a*cos(B-C)/(2S)+b*cos(C-A)/(2S)+c*cos(A-B)/(2S)
=[a*cos(B-C)+b*cos(C-A)+c*cos(A-B)]/(2S)
=R*[sinA*con(B-C)+sinB*cos(C-A)+sinC*cos(A-B)]/S
=(R/S)*[sin(B+C)*con(B-C)+sin(C+A)*cos(C-A)+sin(A+B)*cos(A-B)]
=(R/S)*[sin(2A)+sin(2B)+sin(2C)
=(R/S)*[4sinA*sinB*sinC]
=(R/S)*[S/(2R^2)]=2/R.

In rectangular ABCD, if AP is perpendicular to BC and PD = 3PB, then angle ADB =?

The title is wrong, should be ap vertical BD in P
Because ABCD is a rectangle, AP is perpendicular to BC
So △ ABP is similar to △ DAP
So the square of AP = BP * DP
PD = 3PB
So AP = radical 3PB = radical 3 / 3pd
Because AP is perpendicular to BC
Tanadb = AP / PD = radical 3 / 3
Angle ADB = 30 degrees

In rectangle ABCD, ab = 4, ad = 6, point P is the point on the line where BC side is, and AP = 5, connect PD diagonal AC to e, and calculate AE

When p is between BC, AP = 5, ab = 4, BP = 3, PC = 6-bp = 3
Then AC = 2 * radical 13
Because triangle ade ∽ triangle PEC
So ad: PC = AE: EC = 6:3 = 2:1
So AE = 2 / 3 * AC = 4 / 3 * radical 13

Diagonal of rectangle ABCD AC.BD Intersection point O, P is a point outside the rectangle, and AP vertical PC, Pb vertical PD
No picture

Link Po
Angle APC = 90 degrees
So OA = OP = OC
And because ob = od = OA
So OP = ob = OD
So BPD = 90 degrees

In an equilateral triangle ABC, there is a point P, PA = 6, Pb = 8, PC = 10. Find the area of the triangle ABC

Suppose that the side length of equilateral △ ABC is k, and make high ad on BC side, then BD = K / 2. According to Pythagorean theorem, AD & sup2; = AB & sup2; - BD & sup2; = K & sup2; - K & sup2 / 4 = 3K & sup2 / 4AD = (√ 3) K / 2 area s = 1 / 2 × BC × ad = 1 / 2 × K × (√ 3) K / 2 = (√ 3) K & sup2 / 4, take PA as the edge, and make the first order out of △ ABC

As shown in the figure, P is a point in the regular triangle ABC, and PA = 8, Pb = 6, PC = 10?

It is easy to get that △ ADP is an equilateral triangle. We can get PD = 8, CD = Pb = 6, PC = 10, and easy to get ∠ PDC is RT ∠. We can get ∠ ADC is 150 degree, so ∠ APB is 150 degree. According to the cosine theorem, we can get edge AB & # 178; = Pb & # 178; + PC & # 178; - 2p

If P is a point inside ABC, PA = 3, Pb = 4, PC = 5, then the area of ACP is______ ．

As shown in the figure, △ ACD is obtained by rotating △ ABP 60 ° counterclockwise around point a, then ad = PA = 3, CD = Pb = 4, ∧ APD is equilateral triangle, ∧ PD = PA = 3, ∧ Pd2 + CD2 = 32 + 42 = 25, PC2 = 52 = 25, ∧ Pd2 + CD2 = PC2. According to the inverse theorem of Pythagorean theorem, ∧ PCD is right triangle, ∧ ADC = 150 °, s quadrangle APCD = s ∧ APD + s ∧ PCD = 12 × 3 × (3 × 32) + 12 × 3 × 4 = 934 + 6, and CE ⊥ ad intersection ad is made through point C If the long line is longer than e, then ∠ CDE = 180 ° - ∠ ADC = 180 ° - 150 ° = 30 °. CE = 12CD = 12 × 4 = 2, ｜ s △ ACD = 12ad · CE = 12 × 3 × 2 = 3, ｜ s △ ACP = s quadrilateral apcd-s △ ACD = 934 + 6-3 = 934 + 3

It is known that in △ ABC, ⊙ o with AC side as diameter intersects BC at point D, and take a point E on inferior arc ad to make ∠ EBC = ∠ Dec, extend be, intersect AC at g in turn, intersect ⊙ o at h. verification: AC ⊥ BH

It is proved that the diameter of ⊙ o is ⊙ AC ⊙ ad, ∵ DAC = ⊙ Dec, ∵ EBC = ⊙ Dec, ∵ DAC = ⊙ AC, ∵ ADC = 90 °, ∵ DAC + ⊙ DCA = 90 °, ∵ EBC + ⊙ DCA = 90 °, ∵ BGC = 180 ° - (∵ EBC + ⊙ DCA) = 180 ° - 90 ° = 90 ° and ⊥ AC ⊥ BH

O is the center of the triangle ABC circumscribed circle, connecting Ao, BC, D, Bo, AC, e, Co, AB and F. R is the radius. The proof is: 1 / AD + 1 / be + 1 / CF = 2 / R