Hence PB, opposite y in large triangle: PC, opposite y in small:: PC, opposite x in large : PA, opposite x in small.

Hence, if from a point without a circle two secants be drawn, the product of one secant and its external segment is equal to the product of the other and its external segment. 323. Exercise. - Prove § 322 by drawing A'B and AB'. 324. Def. The projection of a straight line AB, upon another straight line MN, is the portion of MN included between the perpendiculars let fall from the extremities of AB upon the line MN.

In Fig. I A'B' is the projection of AB. In Fig. 2, where one extremity of AB is on MN, AB' is the projection.

PROPOSITION XVIII. THEOREM

325. In any triangle, the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides, minus twice the product of one of these sides and the projection of the other side upon it.

Draw AD perpendicular to CB or CB produced, making CD the projection of AC on CB, and call AB=c; AC=b; BC=a; AD=y; BD=m; CD=n.

In Fig. 1, m=a-n; and in Fig. 2, m=n-a.

c2= a2-2an+n2+y2. (2)

But in the triangle ACD, n2+y2=b2.

Substituting this value in (2),

c2= a2+b2-2an.

SUMMARY: c=m2 + y2= a2-2an + n2 + y2= a2 - 2an + b2.

§ 317

§ 317

Q. E. D. PROPOSITION XIX. THEOREM

326. In an obtuse-angled triangle the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides, plus twice the product of one of these sides and the projection of the other side upon it.

GIVEN-the obtuse-angled triangle ABC with B the obtuse angle.

Draw AD perpendicular to CB produced, making BD the projection of AB on CB, and call AB=c; AC=b; BC=a; AD=y; BD=m; CD = n.

But

And

b2 = n2+y2. (1)
n=a+m.

n2= a2+2am + m2.

Substituting this value of n2 in (I),

b2= a2+2am + m2 +y2. (2)

But in the triangle ABD, m2+y2=c2.

Substituting this value in (2),

b2= a2+c2+2am.

SUMMARY: bo = n2+y2= a2 + 2am + m2 + y2 = a2 + 2am + c2.

§ 317

§ 317

Q. E. D. PROPOSITION XX. THEOREM

327. The bisector of an angle of a triangle divides the opposite side into segments which are proportional to the other two sides.

GIVEN-in the triangle ABC, AD the bisector of the angle A.

TO PROVE

DC AC

DB AB

Draw BM parallel to AD and meeting AC produced

at M.

Then in the triangle BMC, since AD is parallel to BМ,

328. Cor. Conversely, if AD divides BC into two segments which are proportional to the adjacent sides, it bisects the angle BAC.

PROPOSITION XXI. THEOREM

329. The bisector of an exterior angle of a triangle meets the opposite side produced in a point whose distances from the extremities of that side are proportional to the other two sides.

GIVEN-in the triangle ABC, AD the bisector of the exterior an

Draw CM parallel to AD, meeting AB at M.

Then in the triangle BAD, since CM is parallel to AD,