Lemmas in Euclidean Geometry
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 = ∠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 midpoint of QP , the similarity implies that ∠BAM = ∠QAD, from which the result follows.
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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 th...
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