Construction #1: We have two lines, BC and DE (we haven't determined where these points are on these lines), and a point A on BC. Find the two points B and C (on BC on either side of A) so that B is equidistant to A and line DE (BA=BD which is perpendicular to DE) and C is equidistant to A and line DE.

Construction #2: We have two concentric circles. With a given point (the leftmost point in the diagram on the right) as a center, draw a circle that intersects the two concentric circles at two points which are colinear (lie on the same line) with the center of the concentric circles. Further question: How many such circles can we construct?

Construction #3: Inscribe a square in a semicircle. See the diagram.

Construction #4: We have two lines which intersect at some inaccessible point off the screen or blackboard, and we have a point not on either line. Draw the line through the given point that goes through the inaccessible point.

Construction #5: In any angle of a triangle ABC, draw a line DE, with D on AB and E on BC, so that AD = DE = EC. This one seems difficult to me.

Construction #6: Given the lengths of the three altitudes of a triangle, construct the triangle. This seems to be very difficult.

Construction #7: Given the lengths of the three medians of a triangle, construct the triangle. This may take some thought.

Answer #1: This requires a fairly clever insight. The construction: Draw AF perpendicular to DE. Bisect angles BAF and CAF, intersecting line DE at D and E respectively. Draw DB and EC perpendicular to DE and intersecting line BC at B and C respectively. And these are the desired points. The insight: by bisecting angle BAF (similarly CAE) we make three angles equal (congruent), BAD, FAD, and BDA (parallel lines and a transversal), and that makes triangle ADB isosceles, so AB=BD. And that is what we wanted. Staring at the original diagram may not have helped. But drawing a few extra lines, and seeing what the triangles look like, may give you this insight.

Answer #2: As suggested by the above diagram, there are more than one such circle. And that diagram also suggests how to find the colinear points. In the diagram on the right, we see that if we can draw one pair of desired points, then the other point (the center of the circle that we are trying to construct) is on the perpendicular bisector of the line between the desired points. The bisector of the line between the desired points is on one of the two green circles in this diagram, and we can calculate the radii of the two green circles, and construct them. Then we can construct tangents from the other point (the center of the circles we are trying to construct) to these green circles. Draw the lines perpendicular to these tangents, which intersect the original blue circles in the desired points.

Further answer: Now let's answer our further question: how many such circles can we construct? Well, there are four pairs of desired points on the two original concentric circles. But these four pairs of points (two blue and two green in this diagram) lie on only two desired circles (the red circles in the diagram). So the answer is that there are two such circles. We can prove this, using symmetry.

Answer #3: This is fairly easy, but it is educational to try it yourself. Draw the square on the diameter, as shown here. Then draw the two lines from the upper vertices to the center of the circle. These two lines intersect the semicircle at the two points which define two vertices of the square that we want. Complete the construction by drawing the other two sides, perpendicular to the two horizontal sides. Is it a square? Well the two quadrilaterals are dissected into similar triangles, and so are similar, and are both squares.

Actually, your construction could have begun with any square with one side centered on the diameter. The "upper" vertices of this square would always be on the above diagonal lines.

Answer #4: We do it by drawing similar triangles with parallel sides. Using the given point as one vertex, we draw any triangle with two vertices on the two given lines. Elsewhere on one of the lines, we choose a point and draw a triangle with sides parallel to the first triangle. The three lines connecting corresponding vertices all meet in one point, in this case the inaccessible point. The construction works whether the given point is between the given lines or not.

You might want to try proving the all three lines meet in one point. That will help us with the next construction.

Answer #5: We can construct a quadrilateral similar to the desired quadrilateral ADEC. To do this we draw F on CB so CF = AB. We draw a line through F parallel to AC. The circle with center B and radius BA intersects this line at point G, making BG = BA. We can continue drawing the sides of the parallelogram GH and CH (an extension of AC), if we want to, producing quadrilateral ABGH which is similar to the desired quadrilateral (AB = BG = GH, and all of the angles are equal). Then we draw AG which intersects BC at E, and we draw DE parallel to BG, and we have our quadrilateral: AD = DE = EC. It should be fairly easy to show that the two quadrilaterals are similar (see answer #4). We didn't need to draw the green lines in the diagram, it just helped to justify the positioning of point G.

Answer #6: The simplifying insight is that the altitudes are inversely proportional to the sides. We can deduce this from the well-known area formula: A=bh/2, where b is the base. Since we have three altitudes ha, hb, and hc, we have three area formulas: A=aha/2=bhb/2=chc/2. This give us: aha=bhb=chc. If aha is equal to a constant, then a is inversely proportional to ha (making the area equal to 1/2 then aha=1 and a=1/ha), and the same is true of the other sides and their altitudes.

So if we construct a triangle with sides 1/ha, 1/hb, and 1/hc, then this triangle will be similar to the desired triangle. Thumbing through Euclid, there are a number of ways to construct these lengths. One way is shown on the left side of the above diagram. We chose an arbitrary point on the left, and draw three circles with radii equal to the three altitudes.We draw an arbitrary circle that intersects all three of those circles. Each of the lines through the left point and the three intersections also intersects this fourth circle in a second point. If the length of the first line (one of the red lines on the left) is ha, then the length of the line segment from the left point to the second intersection (of that line and the fourth circle) is 1/ha, if we make the distance between the left point and the center of the fourth circle equal to 1. So we easily construct the three 1/h's. Then we construct a triangle with these sides, the blue triangle on the right. Here we find that the shortest altitude is too short. So we then construct a similar triangle with the correct altitude.

Note: Above I used a smaller font size for a subscript, so ha means the altitude on side a. If your browser cannot handle different font sizes, you still should be able to figure out the above two paragraphs.

This is an interesting alternative construction for the above. We construct a triangle from just the three given altitudes (far left). Then a triangle (second figure on the left) made up of just the three altitudes of this triangle would be similar to the desired triangle. Then we construct the triangle, similar to that triangle, with one of the desired altitudes. The large similar triangle.

Construction #7: This is somewhat easier than the previous one. We know that the medians trisect each other. We draw the median AA', and mark the centroid O, 1/3 of the distance from A' to A. We extend AA' to D, so that A'O=A'D. In our finished construction, quadrilateral BOCD will be a parallelogram (the diagonals bisect each other). We draw a circle with center O and radius 2CC'/3 and a circle with center D and radius 2BB'/3. These two circles intersect at point C. We can then locate B in any of several ways, and draw the triangle ABC.