Difficulty: Medium
Correct Answer: A
Explanation:
Introduction / Context:
Orthocentre transformations under triangle permutations are classic: if H is the orthocentre of ABC, then swapping in H with another vertex creates a triangle whose orthocentre is one of the original vertices.
Given Data / Assumptions:
Concept / Approach:
Lines BH and CH are altitudes of ΔABC. In ΔHBC, the altitudes through B and C are along BA and CA respectively (since BH ⟂ AC and CH ⟂ AB). The third altitude of ΔHBC then passes through A, and the three altitudes of ΔHBC meet at A.
Step-by-Step Solution:
In ΔHBC, altitude from B is along BA (perpendicular to HC/AB).Altitude from C is along CA.Their intersection is A; hence A is the orthocentre of ΔHBC.
Verification / Alternative check:
Vector or coordinate proofs reflect the same concurrency.
Why Other Options Are Wrong:
Points N, M, L are generic diagram labels without this universal property.
Common Pitfalls:
Assuming H remains the orthocentre after vertex replacement; it does not in general.
Final Answer:
A
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