2. Right triangle leg ratio
Many relationships in Euclidean geometry do not hold in taxicab geometry. A well-known example is that SAS congruence fails in the taxicab plane; another is that the area of a triangle cannot simply be expressed in the classic 1 2bh (see [2]). Nonetheless, a handful of relationships do remain valid in the taxicab plane. For example, we present the following proposition.
Proposition 1. The ratio between the two legs of a right triangle in the taxicab plane is equal to the ratio between the same two legs in the Euclidean plane.
Proof. Let a and b denote the legs of the right triangle. We denote the taxicab lengths of a and b by aT and bT, and we denote the Euclidean lengths of a and b by aE and bE. Ifa and b are parallel to the coordinate axes, then aT = aE and bT = bE, so clearly aT bT = aE bE . Otherwise, let a have nonzero slope m; this means that b has slope− 1 m. We have the relations
2. Right triangle leg ratioMany relationships in Euclidean geometry do not hold in taxicab geometry. A well-known example is that SAS congruence fails in the taxicab plane; another is that the area of a triangle cannot simply be expressed in the classic 1 2bh (see [2]). Nonetheless, a handful of relationships do remain valid in the taxicab plane. For example, we present the following proposition.Proposition 1. The ratio between the two legs of a right triangle in the taxicab plane is equal to the ratio between the same two legs in the Euclidean plane.Proof. Let a and b denote the legs of the right triangle. We denote the taxicab lengths of a and b by aT and bT, and we denote the Euclidean lengths of a and b by aE and bE. Ifa and b are parallel to the coordinate axes, then aT = aE and bT = bE, so clearly aT bT = aE bE . Otherwise, let a have nonzero slope m; this means that b has slope− 1 m. We have the relations
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