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Junior Balkan Mathematical Olympiad

Geometry Difficulty 4.6 AIME Prove it JBMO

Problem:
Let $ABC$ be a triangle with $\measuredangle B = \measuredangle C = 40^{\circ}$ . The bisector of the $\measuredangle B$ meets $AC$ at the point $D$ . Prove that $\overline{BD} + \overline{DA} = \overline{BC}$ .

Solution

Solution:
Since $\measuredangle BAC = 100^{\circ}$ and $\measuredangle BDC = 120^{\circ}$ we have $\overline{BD} < \overline{BC}$ . Let $E$ be the point on $\overline{BC}$ such that $\overline{BD} = \overline{BE}$ . Then $\measuredangle DEC = 100^{\circ}$ and $\measuredangle EDC = 40^{\circ}$ , hence $\overline{DE} = \overline{EC}$ , and $\measuredangle BAC + \measuredangle DEB = 180^{\circ}$ . So $A, B, E$ and $D$ are concyclic, implying $\overline{AD} = \overline{DE}$ (since $\measuredangle ABD = \measuredangle DBC = 20^{\circ}$ ), which completes the proof.

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