SIMILAR figures are equiangular and have their sides about their equal angles proportional. In the case of triangles, as we have already seen one fact is the consequence of the other, that is to say, if triangles are equiangular their sides about the equal angles are proportional and vice versâ. All regular figures of the same number of sides are similar. In a regular polygon all sides and angles are equal to one another. All circles are similar figures.

In other polygons the angles may be equal, but the sides are not therefore of necessity proportional.

To test whether two polygons are similar, we join all diagonals from one corresponding angular point in each.

If similar, they should now be divided into similar triangles corresponding each to each.

Let ABCD, EFGH be two unequal similar figures, so placed that corresponding sides are parallel, they are now said to be similarly situated, when we join corresponding points A, E;

B, F; D, H; C, G the straight lines obtained all meet at one point called a centre of similitude.

Now reverse the proposition, suppose the figure ABCD to be given and O the centre of similitude, suppose also that H is to be at a given distance from D, or on a given line, then join OA and produce it to meet HE, a parallel to DA, in E, this fixes the position of E. In the same way we can find all the angular points of EFGH. This method is used to inscribe in figures those similar to others, or to circumscribe figures about them.

It is not always necessary to draw the whole of both figures to determine sufficient points.

As an example suppose it is required to inscribe a square in sector PSZ to have two angular points on SZ, one on SP, and one on the arc PZ.

Draw PQ perpendicular to SZ, and PR parallel to it, and equal to PQ. Observe that PR and PQ are two sides of a square which can be completed by drawing a perpendicular from R to SZ produced, and that this square has its sides parallel to the one required where S is a centre of similitude. Join SR meeting the arc PZ in M, then M is one of the angular points required; draw MN perpendicular to SZ, and MK parallel to it; draw KL parallel to PQ or MN, then KMNL, is the square required. Since QPR is half a square, therefore LKM is half a square and the lines MN and LN parallel to KL and KM complete the square. The construction depends upon the fact that KL = KM because PQ = PR.

The centres of similitude of two circles are the points of intersection of their common external and internal tangents (figs. 65 and 66).

To inscribe semicircles in any figures, we find the point at which the semicircle is to touch, then draw a perpendicular there to the tangent, and bisect the right angles so obtained. For example to inscribe a semicircle in an isosceles triangle base TYU, bisect the base at Y, draw the perpendicular YU, bisect the right angles TYU and VYU, the bisectors meet the sides of the triangle in W and X, and WX is the diameter of the semicircle. It is sufficient to remember that the half right angle TYW determines the point W, then draw WX parallel to TU.

(81) Describe a triangle similar to a given one having its sides parallel to those of the given triangle, and an inscribed circle of given radius.

Let ABC be the given triangle. Find O the centre of the inscribed circle of ABC, then O is a 66 centre of similitude." With centre O and the given radius describe a circle. The tangents drawn to this circle parallel to AB, BC, CA form the required triangle DEF. The points of contact are found by drawing perpendiculars to the sides of ABC from O.

If with centre O we describe the inscribed circle of the triangle ABC, the triangles ABC and DEF with the circles form two completely similar figures with O as centre of similitude.

(82) Describe a triangle similar to a given one, having its sides parallel to those of the given triangle and a circumscribing circle of given radius.

Let ABC be the given triangle. Find O the centre of the circumscribing circle of ABC. Join OA, OB, OC, cutting the circle described with centre O and the given radius in D, E, F. DEF is the triangle required.

Compare this construction with 81, as before O the centre of similitude, but now the centre of circumscribing circle not of the inscribed circle.

(83) Describe a circle to touch two given straight lines, and pass through a given point.

Let OA, OB be the given straight lines, P the given point. Bisect the angle AOB by ODF, and take any point D on OF and describe with centre D a circle to touch OA and OB. Let E be one of the points at which OP meets this circle, we get a second solution by taking the other point at which OP meets it.

Draw PF parallel to ED to meet OF in F.

With centre F and distance FP describe the circle required. O, the intersection of the common tangents to the two circles is a centre of similitude; ODE, OFP similar triangles.

(84) To inscribe a series of circles to touch two given straight lines, each circle touching the one before it in the series.

Let OA, OB be the two given straight lines, and suppose the first circle of the series described with given radius to touch OA and OB (71). Draw a tangent CD to the circle at the point where the bisector of the angle AOB meets it. Find the centre of the circle inscribed in the triangle OCD by bisecting the angle ODC, and describe the circle. Repeat the process as often as required.