As shown in the figure, in the known triangle ABC, AB=AC, D is a point in the triangle ABC, and the angle ADB is greater than the angle ADC, which proves that DB is less than DC.

The three situations of D that you say are: on AC; on AB; on BC, which is obviously in contradiction with the meaning that D is triangle ABC. To use junior high school knowledge to solve, I have learned this in junior high school, I don't know if you have heard it, that is: triangle big angle to big side, small angle to small.

On the AC; on the AB; on the BC; on the BC; on the BC; on the BC; on the BC; on the BC; on the BC; on the ABC; on the ABC; on the ABC; on the ABC; on the ABC; on the ABC; on the ABC; on the ABC; on the ABC; on the ABC; on the ABC; on the ABC; on the ABC; on the ABC.

As shown in the figure, in triangle ABC, CD vertical AB is in D, AC=20, C=15, DB=9.{1} Find DC length;{2} Find AB length;{3} Find degree of angle ACB As shown in the figure, in the triangle ABC, vertical AB is in D, AC=20, C=15, DB=9.{1} Find DC length;{2} Find AB length;{3} Find the degree of angle ACB

(1) In RT△BCD, CDB=90°, BC=15, BD=9, CD=BC2BD2=12;(2) in RT△ACD, CDA=90°, AC=20, CD=12, AD=AC2CD2=16;(3) in △ABC, AC=20, BC=15, AB=AD+DB=16+9=25, AC2+BC2=400+225=625=25 2=AB2, ABC is a right triangle.

As shown in the figure, in the square ABCD, the points E and F move on the BC and CD respectively, but the distance AH from A to EF is always equal to the length of AB. During the movement of E and F: (1) Is the size of ∠EAF changed? Please state your reasons. (2) Does the circumference of △ECF change? Please state your reasons. As shown in the figure, in the square ABCD, the points E and F move on the BC and CD respectively, but the distance AH from A to EF is always equal to the length of AB. During the movement of E and F: (1) Is the size of ∠EAF changed? Please state your reasons. (2) Is the circumference of △ECF changed? Please state your reasons.

(1) The size of ∠EAF does not change. The reasons are as follows:
According to the meaning,
AB=AH,∠B=90°,
A H⊥EF,
AHE =90°,
AE=AE,
Rt△BAE≌Rt△HAE (HL),
BAE=∠HAE,
Similarly,△HAF DAF,
HAF=∠DAF,
EAF=∠EAH°∠FAH=1
2∠BAH+1
2∠HAD=1
2(∠BAH°∠HAD)=1
2∠BAD,
BAD =90°,
EAF =45°,
The size of the EAF does not change.
(2) The perimeter of the △ECF does not change for the following reasons:
According to (1), Rt△BAE Rt△HAE,△HAF DAF,
BE=HE, HF=DF,
C△EFC=EF+EC+FC=EB+DF+EC+FC=2BC,
The perimeter of the ECF does not change.

As shown in the figure, in the quadrilateral ABCD, E and F are the midpoint of AB and CD, respectively, and EF=1 2(AD+BC).

Let E and F be the midpoint of BC and CD in the square ABCD respectively, and find the value of tan∠EAF

As shown in the figure, in the spatial quadrilateral ABCD, AB=CD, AB⊥CD, E and F are the midpoint of BC and AD respectively, then the angle formed by EF and AB is ______.

Take the midpoint M of AC and connect EM and FM.
E is the midpoint of BC, EM∥AB and EM=1
2AB;
Similarly: FM//CD and FM=1
2CD,
FEM is the angle formed by straight lines AB and EF,
AB⊥CD, AB=CD, FM=EM, FM⊥EM,
EFM is isosceles right angle triangle, FEM=45°
The answer is 45°.

As shown in the figure, in the known triangle ABC, AB=AC, D is a point in the triangle ABC, and the angle ADB is greater than the angle ADC, which proves that DB is less than DC.