The nature of isosceles triangle; Pythagorean theorem; flat
Properties of row quadrilateral.
Thematic geometry synthesis problem.
Analysis (1) was made by △ABC,
AB=AC= 10 cm, BC= 12 cm, and d is
The midpoint of BC, according to the isosceles triangle triad.
The length of BD and CD can be obtained by the property of one.
And from a=2, △BPQ∽△BDA, the phase is used
The corresponding sides of a triangle are proportional, and you can get
The value of t;
(2) (1) Let p be PE⊥BC in E,
It is easy to prove that the quadrilateral PQCM is a parallelogram.
PB=PQ, divided into segments by parallel lines.
Example theorem, we can get the equation 5 2 t 10 = 1 2.
(6-t) 6. Solve this equation to get the answer;
② First, suppose there is a bisector of point p at ∠ACB.
On the line, the quadrilateral PQCM is parallel to the four sides.
Form, it can be concluded that the quadrilateral PQCM is a diamond.
PB=CQ, PM: BC = AP: Pb, and
The equation set can be obtained, and the t value obtained by solving this equation set is
No, so it may not exist.
Solution: (1)△ABC,
AB=AC= 10cm, BC= 12cm, and d is BC.
The midpoint of,
∴BD=CD= 1 2 years BC =6 cm,
∫a = 2，
∴BP=2tcm,DQ=tcm，
∴BQ=BD-QD=6-t(cm)，
∵△BPQ∽△BDA，
∴BP BD =BQ AB，
That is 2t 6 = 6-t 10,
Solution: t =1813;
(2) (1) The intersection point P is PE⊥BC in E,
∫ Quadrilateral PQCM is a parallelogram,
∴PM∥CQ,PQ∥CM,PQ=CM，
∴PB:AB=CM:AC，
AB = AC，
∴PB=CM，
∴PB=PQ，
∴BE= 1 2 BQ= 1 2 (6-t) cm,
∫a = 5 ^ 2，
∴PB=5 2 Chinese medicine,
∵AD⊥BC，
∴PE∥AD，
∴PB:AB=BE:BD，
That is 5 2t10 =12 (6-t) 6,
Solution: t = 3 2,
∴PQ=PB=5 2 t= 15 4 (cm);
2 does not exist. The reason for this is the following：
∫ Quadrilateral PQCM is a parallelogram,
∴PM∥CQ,PQ∥CM,PQ=CM，
∴PB:AB=CM:AC，
∵AB=AC,∴PB=CM,∴PB=PQ.
If point p is on the bisector of ∠ACB, then
∠PCQ=∠PCM，
∫PM∨CQ，
∴∠PCQ=∠CPM，
∴∠CPM=∠PCM，
∴PM=CM，
∴ Quadrilateral PQCM is a diamond,
∴PQ=CQ，
∴PB=CQ，
PB = atcm，CQ=BD+QD=6+t
(cm),
∴PM=CQ=6+t(cm),AP=AB-
PB= 10-at (cm),
That is at = 6+t 1,
∫PM∨CQ，
∴PM:BC=AP:AB，
∴6+t 12 = 10- in 10,
Simplification: 6at+5t=30②,
Substitute ① into ②, t=-6 1 1,
There is no real number a, so the point p is in ∠ACB.
On the bisector of.