As shown in the figure, in the triangle ABC, BC = 6, e and F are the midpoint of AB and AC respectively, the moving point P is on the ray EF, BP intersects CE at D, bisector of angle CBP intersects CE at Q, when CQ = 1 / 3 * ce, EP + BP=____ .

If the extended BQ crosses EF to o, then Pb = POEP + BP = EO, the triangle EQO is similar to the triangle BCQ, the ratio is 2:1, so the answer is 12

In the triangle ABC, BC > AC, point D is on BC, and DC = AC, the bisector CF of angle ACB intersects ad with F, and point E is the midpoint of ab In the triangle ABC, BC > AC, point D is on BC, DC = AC, bisector CF of angle ACB intersects ad at F, point E is the midpoint of AB, connecting EF. 1: verify EF plane BC2: if the area of quadrilateral BDFE is 6, calculate the area of triangle abd

Because DC = AC, the bisector CF of angle ACB intersects ad with F
So f is the midpoint of AD
And E is the midpoint of ab
So EF is the median line of the triangle abd
EF / / BD is EF / / BC
The area of triangle abd is 6 / (3 / 4) = 8
There is a limit on the number of words

In the triangle ABC, the angle ABC = 90 ° ad bisect angle BAC, DC perpendicular to e, EF to AC to F, EC to ad to o (1) It is proved that triangle deo is equal to triangle DCO; (2) If EF is parallel to BC, will EC bisect the angle def? It is to prove your conclusion

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In a triangle ABC, BC > AC, point D is on BC, and DC = AC, the bisector CF of ∠ ACB intersects ad at F, and E is the midpoint of AB, connecting EF

prove:
BC＞AC,AC=CD
The D is on the line BC
In △ ACD, AC = CD, CF bisection ∠ ACD
∴AC=CD,∠FCA=∠FCD,CF=CF
∴△CFA≌△CFD（SAS）
∴AF=DF
/ / F is the midpoint of AD
∵ e is the midpoint of AB, connecting Fe,
The EF is the median line of △ abd
∴EF‖BC
Get the certificate
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In the triangle ABC, AB is greater than AC, CD bisection angle ACB, ad is vertical DC, f is the midpoint of AC, and extended FD intersects AB at point e. it is proved that EF = 2 / 1BC

First draw the figure;
Prolonging AD and crossing BC to g
Because CD is the bisector of angle ACB
Ad vertical DC
It can be concluded that CD is the middle perpendicular of triangle CAG
Ad = DG
AF = FC
Therefore, DF is the median line of triangular AGC
DF = 1 / 2 GC DF / / GC
That is EF / / BC
Ad = DG
ED is the median line of triangle ABG
The available ed = 1 / 2 BG
And DF = 1 / 2 GC
Then EF = 1 / 2 BC

In the triangle ABC, ad bisection angle BAC, CE is perpendicular to AD and intersected with O, AB is parallel to e, EF is parallel to BC, and EC bisection angle DEF is proved Please, I can't get the picture, please help me to solve the problem.

Because ad bisects ∠ BAC
So ∠ bad = ∠ CAD
Because CE is perpendicular to AD and o
So ∠ AOE = ∠ AOC = 90 degrees
Because Ao = Ao
So △ AOE is all equal to △ AOC
So OE = OC
Because ∠ DOE = ∠ doc = 90 degree od = OD
So △ DOE is all equal to △ doc
So ∠ deo = ∠ dc0
Because EF / / BC
So ∠ DCO = ∠ FeO
So ∠ deo = ∠ FeO
So the EC bisection angle def

In the triangle ABC, the angle ACB = 90 ° D is a point on BC, de ⊥ AB is at point E, and DC = De, ad and CE intersect at point F. it is proved that (1) CF = ef (2) ad ⊥ CE In the triangle ABC, the angle ACB = 90 ° D is a point on BC, de ⊥ AB is at point E, and DC = De, ad and CE intersect at point F. it is proved that (1) CF = ef (2) ad ⊥ CE

∵DC=DE,AD=AD
∴Rt△ACD≌Rt△AED
∴AC=AE,∠CAD=∠EAD
The △ CAE is an isosceles triangle
ν AF ⊥ bisection CE
∴CF=EF,AD⊥CE

As shown in the figure, in the triangle ABC, D is on BC, BD = DC, angle FDE = 90 degrees, F and E are on AB and AC respectively. It is proved that BF + CE > EF

It is proved that: extend ed to point G to make ed = Gd, and connect BG and FG
∵BD＝CD,ED＝FD,∠BDG＝∠CDE
∴△BDG≌△CDE （SAS）
∴BG＝CE
∵BF+BG＞FG
∴BF+CE＞FG
∵∠FDE＝90
ν DF vertical bisection eg
∴EF＝FG
∴BF+CE＞EF

As shown in the figure, the triangle ABC is an isosceles triangle with angle ACB = 90 degrees. The midpoint D of BC is De, perpendicular to AB, and the perpendicular foot is e, connecting CE Find the value of sin angle ace

Make ef ⊥ AC in F, eg ⊥ BC in G
Then BD = BC / 2, CF = eg = BC / 4
∴EF=CG=3BC/4
∴CE=√（CFˇ2+EFˇ2）=（√10/4）BC
∴sin∠ACE=EF/CE=3√10/10≈0.9487

As shown in the figure, in RT △ ABC, ∠ C = 90 °, BC = AC, points D and E are on BC and AC respectively, and BD = CE, M is the midpoint of AB, and △ MDE is isosceles triangle As shown in the figure, in RT △ ABC, ∠ C = 90 °, BC = AC, points D and E are respectively on BC and AC, and BD = CE, M is the midpoint of AB, and △ MDE is isosceles triangle. Please explain the reasons

It is proved that if MC is connected, there is ∠ ECM = ∠ DBM, EC = dB, CM = MB, so △ CEM ≌ △ BDM
Then EM = MD, so △ MDE is an isosceles triangle

As shown in the figure, in the triangle ABC, BC = 6, e and F are the midpoint of AB and AC respectively, the moving point P is on the ray EF, BP intersects CE at D, bisector of angle CBP intersects CE at Q, when CQ = 1 / 3 * ce, EP + BP=____ .