MODEL I

8.

Thus.

Given ARST with a point U inside such that

LUTS

R

PROBLEM I

But

Hence.

Let Af be the midpoint of BC. AM intersects CP at Q. By the symmetry properties. we may have

QBC (QCB = (PCB 3.

LPBQ = LPBCQBC 35"- 30" - 5".

and

S

M

hence

namely, BP bisects ADQ. Morover, since

T

4PBQLABP.

C

(ADP ABC-PBC (180-(BAC)-

namely, QP bisects

follows that