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
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E
B P? cost and freight
Answer? B
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? 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.