14, △ABC and △DBC are right triangles, BC is their hypotenuse, and point P is the midpoint of BC. Connect AD so that PQ in Q is ⊥ ad.

Verification: PQ divides AD equally.

Prove: ∵ According to the given conditions, points A, B, C, D*** circle, BC is the diameter, P is the center of the circle, even PA and PD, then

PA=PD=PB, PQ is an isosceles triangle? The height of the gasket is the center line,

∴AQ=DQ, PQ divides the advertisement equally.

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Ah q? d？ c？

d？

p？

g？ K

?

E

B P？ cost and freight

Answer? B

?

? Title 14? Map number 15

15, e is a point on the side AD of rectangular ABCD, be = ed, p is any point on the diagonal BD, PF⊥BE, PG⊥AD, and the vertical feet are f and g respectively.

Verification: AB = PF+PG

Certificate: PK⊥BC, then G, P, K***, GK∨ab, PG+PK=GK=AB,

∠EBD=∠EDB=∠CBD，

In Rt⊿BPF and Rt⊿BPK, BP is the common line ∠FBP=∠KBP,

∴Rt⊿BPF≌Rt⊿BPK，

∴PF=PK，

∴PF+PG=GK=AB，

∴AB=PF+PG。

Category 26

14, in △ABC, D is the midpoint of AB, E is the midpoint of CD, and the intersection point C is the extension line of CF parallel to AB and AE at point F, connecting BF.

(1) verification: DB = CF

Syndrome: ce = de, cf∨ab, ∴∠ADE=∠FCE (internal angle), and ∠AED=∠FEC.

∴△ADE≌△FCE,∴AD=CF，

AD=BD，∴ BD = CF

(2) If AC = BC, try to judge the shape of the quadrilateral BDCF and prove it.

If AC = BC, CD is the height of isosceles △ AC=BC and the vertical line of the center line, CD⊥CF, FB⊥AB,

The quadrangle DBFC is rectangular,

∴ quadrilateral ABFC is a right-angled trapezoid.

c？ f？

? M’M K F？

? f’C

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? e·O

? E

A b? Answer? B

? d？ e′？ n？

? n′

26. Question 14? MapNo. 16

16. In △ABC, point O is a moving point on the side of AC. When passing through point O, the vertical line MN is parallel to BC. Let the bisector of MN intersecting with ∠BCA be at point E, and the bisector of outer corner intersecting with ∠BCA be at point F (1).

Certificate: MN∨BC, O is the intersection of Mn and AC, ∠ACK is the outer corner of ∠BCA, CF is the angular bisector of ∠ACK, ∠ KCF = ∠ OCF, ∠OFC=∠KCF (inner corner), And prove it. When the O point moves to the AC midpoint, the quadrilateral AECF is rectangular: △ AOE ′△ COF ′, △ AOF ′△ E ′ OC, ∠ E ′ CF = 90.

∠AE′C = 90°

∴ quadrilateral AE'CF' is a rectangle.