Geometry Problem 1621
A Surprising Metric Identity in a 40°–70°–70° Triangle
Problem Statement
In a triangle $ABC$ with $\angle ABC = 40^\circ$, and $\angle ACB = 70^\circ$, let $AH$ be the altitude drawn from vertex $A$ to the side $BC$.
To Prove:
$$AC + AH = BH\sqrt{3}$$
Strategic Hints
- Hint 1 — Look for a $30^\circ\text{--}60^\circ\text{--}90^\circ$ Triangle: The appearance of $\sqrt{3}$ strongly suggests introducing an auxiliary construction that creates a $30^\circ\text{--}60^\circ\text{--}90^\circ$ triangle.
- Hint 2 — Exploit the Isosceles Structure: Since the triangle is isosceles, the altitude from the apex is simultaneously the median and the angle bisector. Use these three roles together.
- Hint 3 — Introduce an Auxiliary Point: Extend one of the altitude rays or construct an auxiliary segment that transforms the desired linear identity into the equality of two geometric lengths.
- Hint 4 — Search for Congruent Triangles: Several $20^\circ$, $40^\circ$, and $70^\circ$ angles naturally appear. Appropriate congruent triangles simplify the expression until the desired identity becomes evident.
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