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As shown in the figure, in triangle ABC, angle ACB = 90 degrees, AE = AC, BC = BD, calculate the degree of angle ECD Ed on ab
From AE = AC, angle AEC = angle ace; from BC = BD, angle BDC = angle BCD. Therefore, angle ACE + angle BCD = angle AEC + angle BDC, that is, angle ACB + angle ECD = angle Dec + angle CDE = 180 ° - angle ECD
The two points E and F on any triangle ABC and BC are equally divided into BC sides, and the intersection line BM of aeaf is connected with G and h, and BG: GH; HM is calculated=
If be = EF = FC, am = MC, MF is a median line of △ AEC, MF ‖ AE, MF = AE / 2;
In △ BMF, because be = EF, GE ∥ MF, BG = GM, or BG; (GH + HM) = 1:1, and MF = 2Ge;
Because MF = AE / 2 = 2Ge, Ge = AE / 4, or Ge = Ag / 3, then MF = (2 / 3) AG,
Because MF ‖ AG, GH: HM = Ag: MF = 3:2,
From BG: (GH + HM) = 1:1 and the former formula, BG: GH: HM = 5:3:2
As shown in the figure. AB = AC, the vertical bisector of AB intersects AC at point D. if ad = BC, (1) Find ∠ B (2) If the point E is on the extension line of BC, and CE = CD, then connect AE, find ∠ CAE
(1) Connecting BD, let ∠ BAC = x °, ∵ the vertical bisector of ∵ AB intersects AC at point D, ? ad = BD, abd = ∠ BAC = x °, ∵ BDC = ∠ BAC + ∠ abd = 2x °, ? ad = BC, ? BD = BC, ab = AC, BCD = ∠ ABC = 2x °, 5x = 180
It is known that in △ ABC, ab = AC, point D and point e are on AB and AC respectively, and ad = AE, be and CD intersect at point O. it is proved that point O is on the vertical bisector of line BC
In △ ADC and △ AEB, it is proved that: in △ ADC and △ AEB, ad = AE 8780 pointo online
It is known that: as shown in the figure, points D and E are on the edge BC of △ ABC, ad = AE, ∠ bad = ∠ CAE. Verification: ab = AC
I don't know if there are three cases of D, e and B in your graph, so the coincidence should not be. E look at this: because ad = AE (known), so ∠ AEC = ∠ ADB (two base angles of isosceles triangle are equal), and because ∠ bad = ∠ CAE (known), so triangle AEC and triangle ADB (ASA) so AB = AC (congruent triangle)
As shown in the figure, in the triangle ABC, ad is perpendicular to BC, BD = CD, and point C is on the vertical bisector of line AE. If AB = 8 and BC = 6, then according to the existing conditions, Can we get the value of de? If so, please give the reason As shown in the following figure, given points e, F and triangle AOB, find a point P so that the distances from point P to both sides of the triangle AOB are equal, and the distances to points E and F are equal. As shown in the figure, in the triangle ABC, ab = AC, O is a point in the triangle ABC, and ob = OC
From ad perpendicular to BC, BD = CD, ab = AC = 8, BD = CD = 3
From point C on the vertical bisector of line AE, AC = CE = 8,
So de = CD + CE = 11
As shown in the figure, in RT △ ABC, ab = AC, ∠ BAC = 90 °, BD bisection ∠ ABC intersects AC at point D, CE ⊥ BD intersects extension line of BD at point E Confirmation: BD = 2ce
It is proved that the extension line of CE to Ba is longer than that of F
Because ∠ Abe = ∠ ACF (equiangular
Equal)
AB=AC
∠BAC=∠CAF=90
So △ abd ≌ △ ACF
So BD = CF
Because BD is both the bisector of angle B and the height of CF side
So △ CBF is an isosceles triangle
CE=1/2CF
BD = CF
So BD = 2ce
As shown in the figure, in RT △ ABC, ab = AC, ∠ BAC = 90 °, 1 = ∠ 2, CE ⊥ BD intersects BD extension line at point e. it is proved that CE = 1 / 2bd
First, make the auxiliary line to extend the extension line of CE to BA at F
Because angle EBF = angle EBC, be = be, angle bef = angle BEC = 90 degrees
So the triangle bef and BEC are congruent
So BC = BF, CE = EF
So CE = 1 / 2 CF
Because the angle abd + ADB = 90 degrees, the angle ECD + CDE = 90 degrees, the angle ADB = CDE
So the angle abd = ECD
Because AB = AC, angle DAB = fac
So the triangle DAB and FAC are congruent
So BD = CF
So CE = 1 / 2 BD
So BD = 2ce
As shown in the figure, in RT △ ABC, ab = AC, ∠ BAC = 90 °, 1 = ∠ 2, CE ⊥ BD is extended to e. it is proved that BD = 2ce
It is proved that: extended CE and Ba intersect at point F, as shown in the figure, ∵ be ⊥ EC, bef = ∠ CEB = 90 °. ? BD bisection ? ABC, ? 1 = ∠ 2, ∵ f = ∠ BCF, ? BF = BC, ∵ be ⊥ CF, ? CE = 12CF, ? ABC, AC = AB, ∵ a = 90 °, CBA = 45 °, f = (180-45) ⊥ 2 = 67.5 °
As shown in the figure, in RT △ ABC, ab = AC, ∠ BAC = 90 °, 1 = ∠ 2, CE ⊥ BD is extended to e. it is proved that BD = 2ce
It is proved that the extended CE and Ba intersect at point F, as shown in the figure,
∵BE⊥EC，
∴∠BEF=∠CEB=90°．
∵ BD bisection ∵ ABC,
∴∠1=∠2，
∴∠F=∠BCF，
∴BF=BC，
∵BE⊥CF，
∴CE=1
2CF，
In ∵ ABC, AC = AB, ∵ a = 90 °,
∴∠CBA=45°，
∴∠F=（180-45）°÷2=67.5°，∠FBE=22.5°，
∴∠ADB=67.5°，
∵ in △ ADB and △ AFC,
∠F=∠ADB
∠BAC=∠FAC
AB=AC ，
∴△ADB≌△AFC（AAS），
∴BD=FC，
∴BD=2CE．
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