As shown in the figure, take sides AC and BC of △ ABC as one side respectively, and make square ACDE and cbfg outside △ ABC. Point P is the midpoint of EF. Prove that the distance from point P to AB is half of ab

As shown in the figure, take sides AC and BC of △ ABC as one side respectively, and make square ACDE and cbfg outside △ ABC. Point P is the midpoint of EF. Prove that the distance from point P to AB is half of ab

Then Er ∥ PQ ∥ FS, ∵ P is the midpoint of EF, ∥ q is the midpoint of RS, ∥ PQ is the median line of trapezoidal EFSR, ∥ PQ = 12 (ER + FS), ∥ AE = AC (equal side length of square), ∥ aer = ∥ cat (equal residual angle of the same angle), ∥ r =

As shown in the figure, in △ ABC, ab = AC, D is a point on the edge of BC, de ⊥ AC is at point E, DF ⊥ AB is at point F, and AE = AF
（1）BD＝CD；
（2）AD⊥BC.

Certification:
（1）∵AB=AC
∴∠B=∠C
∵DE⊥AC,DF⊥AB
∴∠BFD=∠CED=90°
∵DE=DF
∴△BDF≌△CDE
∴BD=CD
（2）
∵BD=CD,AB=AC
⊥ ad ⊥ BC (isosceles triangle with three lines in one)

As shown in the figure, in △ ABC, ab = AC, D is the midpoint of BC, de ⊥ AB, e is the perpendicular foot, DF ⊥ AC, f is the perpendicular foot. Try to explain the truth of de = DF

It is proved that: ∵ AB = AC, D is the midpoint of BC, ∵ bad = ∵ CAD (three lines in one isosceles triangle), ∵ de ⊥ AB, DF ⊥ AC, ∵ de = DF (the distance from the point on the bisector to both sides of the angle is equal)

In the period of Cartesian coordinates, the coordinates of vertex A and B of triangle ABC are (- 1, - 2) (3, - 2) respectively, and vertex C moves on the line y = x + 2 (1) when the triangle ABC
(1) When the area of triangle ABC is 6, try to find the coordinates of point C
(2) When the triangle ABC is an isosceles triangle with ab as the base, the coordinates of point C are obtained

（1）
Obviously, the lines a and B are parallel to the X axis, and ab is 4
If the area of triangle ABC is 6
Then the distance between point C and line AB is 6 / 4x2 = 3
The ordinate of point C is 1 or - 5
Substituting ordinate 1 or - 5 into y = x + 2
The abscissa is - 1 and - 7 respectively
The coordinates of point C are (- 1,1) or (- 7, - 5)
（2）
Obviously, point C is on the vertical bisector of line ab
That is, the abscissa of point C is 1, and the ordinate is undetermined
Substituting abscissa 1 into y = x + 2
The ordinate is 3
The coordinates of point C are (1,3)