# The many ways to construct a triangle II

h

_{c}, l_{c}, m_{c}Assume the triangle ABC is given and draw its circumcircle. Draw also h

_{c}, l_{c}, m_{c}. Extend l_{c}to the intersection with the circle at P and connect M_{c}(I'll write M for simplicity. The same convention applies to L_{c}and H_{c}.) Since P is the midpoint of the arc APB and M is the midpoint of AB, MP is perpendicular to AB and, therefore, is parallel to CH. We conclude that CP (the bisector of C) lies between CH and MP. Finally, the foot of the angle bisector L lies between H and M. This is universally true for every triangle.The center of the circle lies on MP and on the perpendicular bisector of CP. From here we get the following construction:

Form a right triangle CHM. With the center at C sweep an arc of radius l

_{c}. If it does not intersect HM between H and M, the construction is not possible. Otherwise we get a point L. Continue CL beyond L and draw a perpendicular to HM at M. The two lines are bound to intersect at P. Draw a perpendicular bisector of CP till it intersects MP at O. With the radius OP draw a circle with center O. Extend HM both directions to obtain A and B.a, b, A

This is a well known construction and, in principle, the three elements a, b, and A may not define a triangle or may not define it uniquely. When the angle is right the triangle is defined in a unique way.

Draw AC of the given length b and the angle A. The circle of radius a and centered at C may intersect the other leg of A either at two points B', B", at one point B (it will actually touch AB) or not at all. When the point of intersection is unique the angle B is right. If this is known in advance, we may use a different approach.

If B is right, AC is the hypotenuse of the triangle ABC. Its midpoint serves as the circumcenter of the triangle. Draw AC and a circle with diameter AC. From C draw an arc with radius a.

h

_{a}, b, cDraw two parallel lines h

_{a}(units) apart. Pick a point A on one of them and from this point and with the center at A draw arcs of radii a and b till they intersect the other line. The problem may have no solution (e.g., b<h_{a}or b<h_{a}). Otherwise, it may have one solution (b=h_{a}or c=h_{a}). And lastly, it may have two solutions ABC' and ABC" as indicated in the diagram.aa, H

_{b}, H_{c}Triangles BH

_{b}C and BH_{c}C are both right. Therefore, the two given points H_{b}and H_{c}lie on the circle with diameter BC. The perpendicular bisector of the segment H_{b}H_{c}passes through the center M of this circle. This suggests the following construction:Find the point M at the intersection of aa and the perpendicular bisector of H

_{b}H_{c}. Draw a circle centered at M through either H_{b}or H_{c}. Extend BH_{c}and CH_{b}until they intersect at A.Second Construction (Barukh Ziv)

Recalling the mirror property of the orthic triangle, we conclude that angles BH

_{a}H_{b}and CH_{a}H_{b}are equal. This suggests the following construction. Join points H_{b}, H_{c}. Reflect H_{b}in aa, call the image H_{b'}. Then intersection of H_{c}H_{b'}with aa is point H_{a}. Bisector of the angle H_{b}H_{c}H_{a}intersects aa at point C. Through point H_{c}draw a perpendicular to H_{c}C, which intersects aa at B. Finally, the intersection of BH_{c}with CH_{b}is point A.a, h

_{b}, l_{c}Draw an auxiliary triangle BCH

_{b}and the bisector of the angle C. On this bisector, select L_{c}so that CL_{c}= l_{c}. Extend BL_{c}and CH_{b}until they meet at A.h

_{a}, h_{b}, cDraw an auxiliary triangle ABH

_{b}and the circle with radius h_{a}centered at A. Draw a tangent from B to the circle. Extend the tangent and AH_{b}until they meet at C.A second solution starts with a circle, say, ω on c as a diameter. One of the endpoints id naturally A, the other B. Draw circles

C(A, h_ and_{a})C(B, h_ and find their intersections with ω. These will be the feet of the altitudes H_{b})_{a}and H_{b}from A and B, respectively. C lies at the intersection of BH_{a}and AH_{b}.a, b, h

_{a}Let BC = a. Draw a line parallel to BC at distance h

_{a}. Sweep an arc with center C and radius b. This will intersect the line in the sought vertex A.h

_{a}, h_{b}, m_{c}Assume the triangle ABC has been constructed. From M

_{c}drop a line M_{c}H' perpendicular to CB. Then M_{c}H' is half h_{a}. This suggests the following construction:First construct the right triangle CM

_{c}H' with M_{c}H' = h_{a}/2 and hypotenuse CM_{c}= m_{c}. CH' defines the line aa. Similarly, on the other side of CM_{c}find the line bb. Extend CM_{c}to twice its length to get the point D from which draw lines parallel to aa and bb to obtain A and B, respectively.

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