( 1)

∫△ABC is an isosceles triangle

D is the midpoint of BC, according to the properties of isosceles triangle:

AD⊥BC

According to the meaning of the question, PQB < 90

So when △BPQ∽△BDA,

∠BPQ =∠ ADB =90

BQ/AB=BP/BD

BQ = BD-DQ = 6t

(6-t)/ 10=2t/6

t= 18/ 13

(2)

①

PQCM is a parallelogram.

∴PQ//MC

PM//QC

Therefore,

△BPQ∽△BAC

BP/BA=PQ/AC=BQ/BC

BP=5t/2

BQ=6-t

Therefore:

(5t/2)/ 10=(6-t)/ 12

t=3/2

Therefore:

PQ = AC *(3/8)= 10×(3/8)= 15/4

②

When p is on the bisector of ∠ACB,

AC/CB=AP/PB

AP= 10-at

∴

10/ 12 =( 10-at)/at

at=60/ 1 1

t=60/ 1 1a

At this point:

DQ=60/ 1 1a

BQ = 6-60/ 1 1a & gt； 0

Namely:

6 & gt60/ 1 1a

a & gt 10/ 1 1

Therefore:

Such a point P exists.