As shown in the figure, in the isosceles triangle ABC, ab = AC, point D is a point on BC, de ∥ AC intersects AB at point E, DF ∥ AB intersects AC at point F, ① As shown in Figure 1, when point D is on BC, there is de + DF = AB, please explain the reason. ② as shown in Figure 2, when point D is on the extension line of BC, please refer to figure 1 to draw the correct figure, write down the relationship between De, DF and AB, and write the proof process,

As shown in the figure, in the isosceles triangle ABC, ab = AC, point D is a point on BC, de ∥ AC intersects AB at point E, DF ∥ AB intersects AC at point F, ① As shown in Figure 1, when point D is on BC, there is de + DF = AB, please explain the reason. ② as shown in Figure 2, when point D is on the extension line of BC, please refer to figure 1 to draw the correct figure, write down the relationship between De, DF and AB, and write the proof process,

One
When point D is on BC
∵DE∥AC
∴∠EDB=∠ACB
∵ in the isosceles triangle ABC, ab = AC
∴∠ABC=∠ACB
∴∠EDB=∠ABC
∴DE=BE
∵ DF ∥ AB intersects AC at point F
The AEDF is a parallelogram
∴DF=AE
∴DE+DF=AB
Two
When point D is on the extension line of BC,
De ∥ AC intersects AB extension at point E
DF ∥ AB intersects AC extension at point F,
Then de-df = ab
∵DE∥AC
∴∠EDB=∠ACB
∵ in the isosceles triangle ABC, ab = AC
∴∠ABC=∠ACB
∴∠EDB=∠ABC
∴DE=BE
∵DF∥AB
The AEDF is a parallelogram
∴DF=AE
∴DE-DF=AB

In isosceles △ ABC, ab = AC, D is a moving point on BC, de ∥ AC, DF ∥ AB, which intersect AB at e and AC at f respectively, then does de + DF change with the change of D point? Please state the reasons

No change. The reasons are as follows:
∵DE∥AC，DF∥AB
The quadrilateral AEDF is a parallelogram
/ / DF = AE (opposite sides of parallelogram are equal)
And ∵ AB = AC
Ψ B = ∠ C (equilateral and equal angle)
∵DE∥AC
∴∠EDB=∠C
﹤ EDB = ∠ B (equivalent substitution)
ν de = EB (equiangular to equilateral)
∴DE+DF=AE+EB=AB．

In isosceles △ ABC, ab = AC, ab = 5cm, D is any point on BC edge, DF / / AC, de / / AB, calculate the circumference of parallelogram aefd

∵ the triangle ABC is an isosceles triangle, ab = AC = 5, and ∵ DF / / AC, de / / AB, it is easy to prove that the triangle FBD and the triangle EDC are isosceles triangles.

As shown in the figure, D is the midpoint on the BC side of △ ABC, de ⊥ AC, DF ⊥ AB, the perpendicular feet are e and f respectively, and BF = CE. It is proved that ⊥ ABC is an isosceles triangle Verification: 1. △ ABC is isosceles triangle 2. When ∠ a = 90 °, try to judge what kind of quadrilateral afde is and prove your conclusion

1) It is proved that in △ BFD and △ CED, BD = CD, be = CE, ∠ DFB = ∠ Dec = 90 degrees
Then: △ BFD and △ CED are congruent
Then ∠ B = ∠ C
So delta ABC is an isosceles triangle
2) The quadrilateral afde is square
It is proved that when ⊥ a = 90 °, de ⊥ AC, DF ⊥ ab
Then the quadrilateral afde is rectangular
(1) It has been proved that △ ABC is an isosceles triangle
Then AB = AC, BF = CE, AF = AE
So the quad afde is square

As shown in the figure, we know that in △ ABC, D is the midpoint of BC, de ⊥ AB, DF ⊥ AC, and the perpendicular feet are e and f respectively, and de = DF. Try to explain that △ ABC is an isosceles triangle

To prove isosceles, we only need to prove AC = ab
Connecting ad D is the midpoint of BC, so de = DF, ad = da
From de ⊥ AB DF ⊥ AC ⊥ it can be concluded that ⊥ AED = ∠ AFD = 90 °
So △ ade ≌ △ ADF
AE = AF
Re prove be = CF
(D is the midpoint BD = CD, de = DF ∠ bed = ∠ CFD = 90 ° to prove △ bed ≌ △ CFD)
So ABC + AE is a triangle, which is a triangle

As shown in the figure, in the isosceles triangle ACB, AC = BC = 5, ab = 8, D is the moving point on the bottom AB (not coincident with points a and b), de ⊥ AC, DF ⊥ BC, and the perpendicular feet are e and f respectively, then de + DF=______ .

Connect CD and pass through point C as the high CG on the bottom edge ab,
∵AC=BC=5，AB=8，
∴BG=4，CG=
BC2−BG2=
52−42=3，
∵S△ABC=S△ACD+S△DCB，
∴AB•CG=AC•DE+BC•DF，
∵AC=BC，
∴8×3=5×（DE+DF）
∴DE+DF=4.8．
So the answer is: 4.8

As shown in the figure, in △ ABC, Ba = BC, point D is a point on the extension line of AB, DF ⊥ AC at point F is called BC at point e. it is proved that △ DBE is an isosceles triangle

Ba = BC, so ∠ a = ∠ C,
Because ∠ a + ∠ d = 90 °, C + ∠ CEF = 90 °,
So ∠ d = ∠ CEF, because ∠ CEF = ∠ bed,
So ∠ d = ∠ bed
So be = BD
So delta DBE is an isosceles triangle

In the isosceles triangle ABC, ab = AC = 10, BC = 16, D is the midpoint of BC side. Find the distance between point D and AB, AC

Connecting ad, i.e., ad is the height of isosceles triangle ABC, abd is a right triangle, BC = 16, D is the midpoint of BC side, so BD = 8 8? + Ad 2 = 10? Ad = 6S △ ABC = 16 × 6 △ 2 = 48 ∵ △ abd and △ ACD are congruent triangles, and △ abd = 48 △ 2 = 24D, and the distance from AB is also the height of △ abd

As shown in the figure, in △ ABC, ab = AC, D is the midpoint of BC. It is proved that the distances from D to AB and AC are equal

It is proved that the ad is connected with D as de ⊥ AB to e, DF ⊥ AC to F,
∵ D is the midpoint of BC,
∴BD=CD，
∵ in △ abd and △ ACD
AB＝AC
AD＝AD
BD＝DC ，
∴△ABD≌△ACD（SSS）．
∴∠BAD=∠CAD，
∵DE⊥AB，DF⊥AC，
∴DE=DF，
That is, the distance from D to AB and AC is equal

As shown in the figure, in the isosceles triangle ABC, given AB = AC, ∠ a = 30 ° and the vertical bisector of AB intersects AC at D, then the degree of ∠ CBD is______ °．

∵AB=AC，∠A=30°，
∴∠ABC=∠ACB=75°，
∵ the vertical bisector of AB intersects AC with D,
∴AD=BD，
∴∠A=∠ABD=30°，
∴∠BDC=60°，
∴∠CBD=180°-75°-60°=45°．
Therefore, fill in 45