M is one point on the side of the triangle ABC BC

p is half circumference:

p=(a+b+c)/2

Suppose that the radius of the inscribed circle is r, then the area of the triangle is s=1 2r*(a+b+c), because the radius of the inscribed circle is perpendicular to the three sides.

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All of the above are correct! In fact, to put it bluntly, the area method is used.

The radius of the inscribed circle can be found using the area method.

By the title, mc=mb=5, am=3, to do Zheng Yuan quantity mc=- vector mb

Vector ab = vector mb - vector ma, vector ac = vector mc - vector ma, vector ab dot product vector ac= (vector mb - vector pure state ma) * (vector mc - vector ma) = (vector mb - vector ma) * (vector mb - vector ma).

Square of the vector MA - Square of the vector MB = 3 2-5 2 = -16

Copying the wrong question. A point satisfying the known conditions can be any point on the circumference of a circle centered on m and radius 3, and ab*bc is not a fixed value.

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Proof : Intercept EA=MC on AB and connect EM to obtain AEM 1=180°- AMB- AMN, 2=180°- AMB- B, AMN= B=60°, 1= 2

and cn bisect acp, 4= acp=60° mcn= 3+ 4=120°....①

and ba=bc, ea=mc, ba-ea=bc-mc, i.e. be=bm bem is an equilateral triangle 6=60°

By mcn= 5

In AEM and MCN, 1= 2 AE=MC, MCN= 51= 2 AE=MC, MCN= 5

1=∠2．ae=mc，∠mcn=∠5

aem≌△mcn （asa）．∴am=mn．<>

Proof : Intercept EA=MC on AB and connect EM to obtain AEM 1=180°- AMB- AMN, 2=180°- AMB- B, AMN= B=60°, 1= 2

and cn bisect acp, 4= 12 acp=60° mcn= 3+ 4=120°....①

and ba=bc, ea=mc, ba-ea=bc-mc, i.e. be=bm bem is an equilateral triangle 6=60°

By mcn= 5

In AEM and MCN, 1= 2 AE=MC, MCN= 5 AEM MCN (ASA) AM=MN

Two methods. Proof: Intercept EA=MC on AB, connect EM, and get AEM

1=180°-∠amb-∠amn，∠2=180°-∠amb-∠b，∠amn=∠b=60°，∴1=∠2．

and cn bisect acp, 4= acp=60° mcn= 3+ 4=120°....①

and ba=bc, ea=mc, ba-ea=bc-mc, i.e. be=bm

BEM is an equilateral triangle 6=60°

By mcn= 5

In AEM and MCN, 1= 2 AE=MC, MCN= 5

1=∠2．ae=mc，∠mcn=∠5

1=∠2．ae=mc，∠mcn=∠5

AEM MCN (ASA) AM=MN connects AN.

Because the angle AMN = 60 degrees, therefore, the angle AMB + angle CMN = 120 degrees;

Because the angle abc = 60 degrees, therefore, the angle bam + angle amb = 120 degrees;

So, angular bam = angular cmn.

Because the angle ACP=120 degrees, CN bisects the angle ACP, so, the angle ACN=60 degrees, so, the angle ACN=angle AMN, so, A, M, C, N four points are round, therefore, the angle CMN= angle can.

And because the angle ABC=angle ACN=60 degrees, AB=AC, the triangle ABM congruent triangle ACN(ASA).

So, am=an, and because the angle amn=60 degrees, the triangle amn is an equilateral triangle, so, am=mn.

Connect AN. Because the angle AMN = 60 degrees, therefore, the angle AMB + angle CMN = 120 degrees;

Because the angle abc = 60 degrees, therefore, the angle bam + angle amb = 120 degrees;

So, angular bam = angular cmn.

Because the angle ACP=120 degrees, CN bisects the angle ACP, so, the angle ACN=60 degrees, so, the angle ACN=angle AMN, so, A, M, C, N four points are round, therefore, the angle CMN= angle can.

And because the angle ABC=angle ACN=60 degrees, AB=AC, the triangle ABM congruent triangle ACN(ASA).

So, am=an, and because the angle amn=60 degrees, the triangle amn is an equilateral triangle, so, am=mn.

Find the AF perpendicular bisect cd].

Proof: CD bisects EDF

edc=∠fdc

de//bc

edc =∠dcf

No fdc = dcf

df =cf

and ad=ac, af=af

ADF Fortune Circle ACF (SSS).

daf=∠caf

ADC is an isosceles triangle.

AF bisects DAC, according to the three-in-one.

AF Vertical Macro Bisect CD