Lemmas in Euclidean Geometry 1
نویسنده
چکیده
Second proof. Let O be the circumcenter of ABC and let ω be the circle centered at D with radius DB. Let lines AB and AC meet ω at P and Q, respectively. Since ∠PBQ = ∠DQC + ∠BAC = 1 2(∠BDC + ∠DOC) = 90 ◦, we see that PQ is a diameter of ω and hence passes through D. Since ∠ABC = ∠AQP and ∠ACB = ∠APQ, we see that triangles ABC and AQP are similar. If M is the midpoint of BC, noting that D is the midpoint of QP , the similarity implies that ∠BAM = ∠QAD, from which the result follows.
منابع مشابه
Some Lemmas to Hopefully Enable Search Methods to Find Short and Human Readable Proofs for Incidence Theorems of Projective Geometry
Search methods provide short and human readable proofs, i.e. with few algebra, of most of the theorems of the Euclidean plane. They are less succesful and convincing for incidence theorems of projective geometry, which has received less attention up to now. This is due to the fact that basic notions, like angles and distances, which are relevant for Euclidean geometry, are no more relevant for ...
متن کاملSpatial Analysis in curved spaces with Non-Euclidean Geometry
The ultimate goal of spatial information, both as part of technology and as science, is to answer questions and issues related to space, place, and location. Therefore, geometry is widely used for description, storage, and analysis. Undoubtedly, one of the most essential features of spatial information is geometric features, and one of the most obvious types of analysis is the geometric type an...
متن کاملAutomated theorem proving for elementary geometry
Thesis: Automated theorem proving for elementary geometry. This study analyses automated proofs of theorems from Euclidean Elements , book VI, using the area method. The theorems we will be discussing concern Euclidean field theory about equality of non-congruent figures and similarity of the figures [1]. The proofs are generated by the program WinGCLC. My proposed hypothesizes: 1. The way of m...
متن کاملLemmas in Euclidean Geometry
Second proof. Let O be the circumcenter of ABC and let ω be the circle centered at D with radius DB. Let lines AB and AC meet ω at P and Q, respectively. Since ∠PBQ = ∠BQC+∠BAC = 1 2 (∠BDC + ∠DOC) = 90, we see that PQ is a diameter of ω and hence passes through D. Since ∠ABC = ∠AQP and ∠ACB = ∠APQ, we see that triangles ABC and AQP are similar. If M is the midpoint of BC, noting that D is the m...
متن کاملA Combination of Nonstandard Analysis and Geometry Theorem Proving, with Application to Newton's Principia
The theorem prover Isabelle is used to formalise and reproduce some of the styles of reasoning used by Newton in his Principia. The Principia’s reasoning is resolutely geometric in nature but contains “infinitesimal” elements and the presence of motion that take it beyond the traditional boundaries of Euclidean Geometry. These present difficulties that prevent Newton’s proofs from being mechani...
متن کامل