Triangle proof questions, math triangle proof questions

Extending the extension line of BE AC at N, bisecting BAC and BE perpendicular to AD by AD, we can get the congruence of triangle ABE and triangle ANE, so E is the midpoint of Bn and M is the midpoint of BC to get EM is the median line of the triangle BNC, so EM 1 2CN 1 2 (An AC) 1 2 (AB AC).

Proof: Extend BE and AC to F

ad bisects bac, be ad

bae = fae aoeb = aef = 90 degrees ae = ae abe afe

af=ab be=ef

m is the midpoint of bc.

EM is the median of the BFC.

em=1/2*fc

fc=af-ac af=ab

em=1/2*（ab-ac）

Extend the intersection of ac and be at point f

AD bisects the BAC

Be the vertical AD extension line at E i.e. AE BF

afe≌△abe

af=ab,fe=be

fc=af-ac=ab-ac

E is the midpoint of BF and M is the midpoint of BC.

me=1/2fc=1/2(ab-ac)

The extension cable extending AC and BE intersects F

Then AE bisects BAF and perpendicular BF then AB=AF and E is the midpoint of BF, CF=AB-AC

m is the midpoint of bc em=1 2cf=1 2(ab-ac).

Proof that the parallel lines of the crossing point F as BC cross AC at HLet EF intersect with the x-axis at d;

afg= b, c= g, b= c obtained by ab=ac; ∴∠afg=∠g;

i.e. af=gc; and ab=ac;

bf=cg;bf=ce;

cg=ce;CD is the median line of EFG.

i.e. EF is bisected by the x-axis.

No way. Unless ab=ac, you are a small question, the general question is not, and the conditions are missing.

If ab=ac, you can do an x-axis parallel line with a point by f, and then it is well proved.

eg=1/2bc, dg=1/2bc

Then: eg=dg

The triangle EGD is an isosceles triangle.

F is the midpoint of ED, so: FG is perpendicular to ED (3-in-1).

Let the degree closure number of the three inner angles be x, 2x, 3x respectively, according to the sum of the inner angles of the triangle is 180 degrees, that is, x+2x+3x=180 is solved to obtain x=30, so the degrees of the three inner beam companion angles are 30, 60, and 90 degrees respectively.