THEOREM 4.

Equivalent triangles on equal bases (or, on the same

base) are of the same altitude.

Data ABC, DEF are equivalent triangles on equal bases BC, EF; AL, DM are the altitudes of the two triangles.

COR. 1. If two equivalent triangles stand upon the same base, and on the same side of it, the line joining their vertices is parallel to the base.

(For the altitudes of the triangles are equal.)

COR. 2. If two equivalent triangles stand upon equal bases in the same straight line, and on the same side of it, the line joining their vertices is parallel to their

bases.

Ex. 993.

What is the converse of the above Theorem?

R

Ex. 994. DE are the mid-points of the sides AB, AC of a triangle ABC; prove that DE is parallel to BC. (Join DC, EB.)

Ex. 995. In fig. 180 ▲ PXQ= ▲ RXS; prove that PR is parallel to QS.

Ex. 996. In fig. 181 ▲ AEB= AADC; prove that DE is parallel to BC.

X

fig. 180.

E

B

fig. 181.

THEOREM 5.

If a triangle and a parallelogram stand on the same base and between the same parallels, the area of the triangle is half that of the parallelogram.

Data

A EBC and ||ogram ABCD stand on the same base BC and between the same parallels BC, AE.

Ex. 997.

Construct a rectangle equal to a given triangle.

Ex. 998. F, E are the mid-points of the sides AD, BC of a parallelogram ABCD; P is in any point in FE. Prove that A APB ABCD.

Ex. 999. P, Q are any points upon adjacent sides AB, BC of a parallelogram ABCD; prove that ▲ CDP = A ADQ.

Ex. 1000. AB, CD are parallel sides of a trapezium ABCD; E is the mid-point of AD; prove that ▲ BEC = trapezium. (Through E draw line parallel to BC.)

Ex. 1001.

O is a point inside a parallelogram ABCD; prove that
AOAB+ A OCD=ABCD.

MISCELLANEOUS EXERCISES ON AREA.

Ex. 1002. Find the area of a triangle whose sides are

Ex. 1003.

of a rectangle with the same sides.

The area of a parallelogram of angle 30° is half the area

Ex. 1004*. O is any point on the diagonal BD of a parallelogram ABCD. EOF, GOH are parallel to AB, BC respectively. Prove that parallelogram AO=parallelogram CO.

D

Ex. 1005. Any straight line drawn through the centre of a parallelogram (i.e. through the intersection of the diagonals) bisects the parallelogram.

Ex. 1006. Divide a parallelogram into three equal parallelograms.

Ex. 1007. Bisect a parallelogram by a straight line drawn perpendicular to a side.

* This exercise appears in old books on Geometry as a proposition, and was used by Euclid in the proof of later propositions. It was enunciated as follows: "The complements of the parallelograms which are about the

diagonal of any parallelogram are equal."

Ex. 1008. E is any point on the diagonal AC of a parallelogram ABCD. Prove that ▲ ABEAADE. (Draw the other diagonal.)

Ex. 1009. Produce the median BD of a triangle ABC to E, making DE=DB. Prove that A EBC= AABC.

Ex. 1010. P, Q are the mid-points of the sides BC, AD of the trapezium ABCD; EPF, GQH are drawn perpendicular to the base. Prove that trapezium=rectangle GF. (See fig. 184.)

Ex. 1011. L, M are the mid-points of the parallel sides AB, CD of a trapezium ABCD. Prove that LM bisects the trapezium.

Ex. 1012. In Ex. 1011 O is the mid-point of LM; prove that any line through O which cuts AB, CD (not produced) bisects the trapezium.

Ex. 1013. Prove that the area of the parallelogram formed by joining the mid-points of the sides of any quadrilateral ABCD (see Ex. 736) is half the area of the quadrilateral.

Ex. 1014. The medians BD, CE of ▲ ABC intersect at G; prove that quadrilateral ADGE=A BGC. (Add to each a certain triangle.)

THE THEOREM OF PYTHAGORAS.

Fig. 185 represents an isosceles right-angled triangle with squares described upon each of the sides. The dotted lines divide up the squares into right-angled triangles, each of which is obviously equal to the original triangle. This sub-division shows that the square on the hypotenuse of the above right-angled triangle is equal to the sum of the squares on the sides containing the right angle. (A tiled pavement often shows this fact very clearly.)

fig. 185.