In the triangle ABC, ab = BC, the point D E F is the midpoint on the side of BC AC AB respectively

In the triangle ABC, ab = BC, the point D E F is the midpoint on the side of BC AC AB respectively

De = EF = BF = BD = BC / 2 = BA / 2
The four sides are equal and parallel to each other, and the quadrilateral bdef is a diamond

EC is perpendicular to plane ABC, BD is parallel to CE, and CE = 2 BD.BC=BA Ask if there is a point m on EA so that DM is perpendicular to ECA?
Find whether there is m

Take the midpoint P of AC, connect PM, then PM / / CE, connect Pb, because CE and BD are perpendicular to plane ABC, so CE / / BD, that is PM / / BD, so triangle PMB and triangle BDM are on the same plane, so triangle should be positive

It is known that both sides AB and AC of plane α and △ ABC intersect at D and e respectively, and ad: DB = AE: EC. The BC ‖ plane α is proved

It is proved that: connecting De, ∵ ad: DB = AE: EC, ∵ de ∥ BC, ∵ de ⊂ plane α, BC ⊄ plane α, ∥ BC ⊉ plane α

As shown in the figure, ad is the bisector of ∠ BAC, de ⊥ AB is in E, DF ⊥ AC is in F, and DB = DC

Ad is

In the triangle ABC, the angle ACB = 90 degrees, D is on BC, De is perpendicular to AB, AB intersects e, ed = DC, de extends to F, DF = dB, and ad ‖ FB is obtained

It is easy to prove that △ ADC ≌ △ ade, 2 ∠ ade + BDE = 180 degrees
The △ BFD is isosceles triangle, 2 ∠ F + ∠ BDE = 180 degrees
Therefore, ADE = F
So ad ‖ FB

As shown in the figure, ad is the bisector of ∠ BAC, de ⊥ AB, e for perpendicular foot, DF ⊥ AC, f for perpendicular foot, and BD = CD

It is proved that: ∵ ad is the bisector of ∠ BAC, de ⊥ AB, DF ⊥ AC, ∵ de = DF, and ∵ BD = CD, ≌ RT △ DBE ≌ RT △ DCF (HL). ∵ be = CF

As shown in the figure, BF is parallel to the diagonal AC of square ABCD, point E is on BF, and AE = AC, CF ‖ AE, then the degree of ∠ BCF is______ ．

Through point a, make the extension line of Ao ⊥ FB at point O, connect BD, intersect AC at point Q, ∵ quadrilateral ABCD is square, ∵ BQ ⊥ AC ∵ BF ∥ AC, ∵ Ao ∥ BQ and ∵ qAB = ∵ QBA = 45 °, Ao = BQ = AQ = 12ac, ∵ AE = AC, ∵ Ao = 12ae, ∵ AEO = 30 °, ∵ BF ∥ AC, ∵ CAE = ∵ AEO = 30 °, ∵ BF

It is known that E and F are two points on AB, AE = BF, AC / / DB, and AC = dB

Certification:
∵AC//DB
∴∠A＝∠B
∵AF＝AE+EF,BE＝BF+EF,AE＝BF
∴AF＝BE
∵AC＝DB
∴△ACF≌△BDE （SAS）
∴CF＝DE

Ad is the bisector of the angle BAC, De is perpendicular to AB, the extension of AB is perpendicular to e, DF is perpendicular to AC is perpendicular to F, and DB = DC

∵ ad is the angular bisector of ∠ BAC, de ⊥ AB, DF ⊥ AC
Ψ DF = de (the distance from the point on the bisector to both sides of the angle is equal)
In RT △ bed and RT △ CFD
DF=DE
DB=DC
∴Rt△BED≌Rt△CFD（HL)
Be = CF (the corresponding sides in congruent triangles are equal)

As shown in the figure, be = CF, the extension of de ⊥ AB is at point E, DF ⊥ AC is at point F, and DB = DC

It is proved that: the extension line of ∵ de ⊥ AB is at point E, DF ⊥ AC is at point F, ∵ bed = ∠ CFD, ∵ BDE and △ CDF are right triangles, ∵ be = cfbd = CD, ≌ RT △ BDE ≌ RT △ CDF,