PROPOSITION XI. THEOREM. 338. In any triangle, if a medial line be drawn from the rertex to the base :

I. The sum of the squares on the two sides is equivalent to twice the square on half the base, increased by twice the square on the medial line;

II. The difference of the squares on the two sides is equivalent to twice the product of the base by the projection of the medial line upon the base.

BC

ID
In the triangle A B C let AM be the medial line and

M D the projection of A M upon the base B C.
Also let A B be greater than A C.
We are to prove

I. A B’ + A C = 2 B 1 + 2 A M°.

II. A B’ – A C = 2 BC X MD. Since A B > AC, the L A M B will be obtuse and the LAMC will be acute.

Then A Bo = B M + A 12 + 2 B M X MD, § 336 (in any obtuse A the square on the side opposite the obtuse Z is equivalent to

the sum of the squares on the other two sides increased by twice the product of one of those sides and the projection of the other on that side);

and À C2 = M (2 + AM – 2 MC XMD, $ 335 in any A the square on the side opposite an acute Lis equivalent to the sum

of the squares on the other two sides, diminished by twice the product
of one of those sides and the projection of the other upon that side).
Add these two equalities, and observe that B M = MC.
Then A B’ + C2 = 2 B MP + 2 A M?.
Subtract the second equality from the first.
Then A B – À C2 = 2 B C X M D.

Q. E. D.

PROPOSITION XII. THEOREM. 339. The sum of the squares on the four sides of any quadrilateral is equivalent to the sum of the squares on the diagonals together with four times the square of the line joining the middle points of the diagonals.

B

In the quadrilateral A B C D, let the diagonals be AC

and BD, and FE the line joining the middle
points of the diagonals.
We are to prove
A B + B C + C D? + DĀ = A C2 + B DP + 4 Ê 752.
Draw B E and D E.

Now AB + BC" = 2 (49)* + 2 BE”,

$ 338

(the sum of the squares on the two sides of a A is equivalent to twice the square on half the base increased by twice the square on the medial line to the base), and CD + DĀ = 2

2 DE? $ 338 Adding these two equalities, A B + B C + CD+ DĀ = 4 + 2 (B E+ D E).

But BEP + DE= 2 )2 + 2 È F2, § 338 (the sum of the squares on the two sides of a A is equivalent to twice the square on half the base increased by twice the square on the medial line to the base).

Substitute in the above equality for (BE? + D E%) its equivalent;

then AP® + BC* + CD' + DA’=4 (49) +4 (BD) + 4 EF?

= AC? + BD? + 4 E F?

Q. E. D. 340. COROLLARY. The sum of the squares on the four sides of a parallelogram is equivalent to the sum of the squares on the diagonals.

PROPOSITION XIII. THEOREM.

341. Two triangles having an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles.

Let the triangles A B C and A D E have the common

angle A.
We are to prove

A ABC ABXAC
AADE – ADX A ER

Draw B E.

§ 326

A ABC AC
Now

A ABE – AE'
(A having the same altitude are to each other as their bases).

§ 326

Also

A ABE AB

AADE – A D'
(A having the same altitude are to each other as their bases).

Multiply these equalities;

A A B C _ then

A A DE

A B X 4 C
ADX A E

Q. E. D.

PROPOSITION XIV. THEOREM. 342. Similar triangles are to each other as the squares on their homologous sides.

A

AL

B

Let the two triangles be AC B and A'C' B'.

A ACB Ā B?
We are to prove

A ACB

B' A' B12 Draw the perpendiculars C O and C" O'. Then

A ACB A B XCO ABCO

A A'C' B' A' B'X CO = AR^ TI I § 326 (two A are to each other as the products of their bases by their altitudes). But A B co

§ 297 A' B coi (the homologous altitudes of similar s have the same ratio as their homolo

gous bascs). Substitute, in the above equality, for

Y, for ci o its equal a pii

its equal AB
A ACB
then

A

B A
then

B _ A B2
A A'C' B' = A' B'. * A' B Biz

Q. E. D.

PROPOSITION XV. THEOREM.

343. Two similar polygons are to each other as the squares on any two homologous sides.

В

TE
F
Let the two similar polygons be A B C, etc., and
A' B' C', etc.

A B C, etc. ABC
We are to prove -

A' B' C', etc. A B2
From the homologous vertices A and A' draw diagonals.

ABBC CD
Now

A B = B' C = C Di, etc.,
(similar polygons have their homologous sides proportional);

A B2 BO2 CO2
.. by squaring,

'AT B12 B1 CH2 CDi2' The A ABC, AC D, etc., are respectively similar to A'B'C', A' C'D', etc.,

$ 294 (tuo similar polygons are composed of the same number of similar to each

other and similarly placed).

. Δ Α Β
A B2

§ 342 * A A'B'C' = A' B' (similar S are to each other as the squares on their homologous sides),

A ACD C D and

$ 342 A A'C' D' C' Dj2"