Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Prove it North Macedonia

In the triangle $ABC$ on the side $AC$ a point $D$ is chosen such that $\angle BDC = \angle ABC$ . If it is known that $\overline{AD} = 7\text{cm}$ and $\overline{DC} = 9\text{cm}$ , calculate the length of $BC$ and the ratio $\overline{BD}:\overline{BA}$ .

Solution

Because $\angle BDC = \angle ABC = \beta$ and $\angle DCB = \angle BCA = \gamma$ we have that $\triangle BDC \sim \triangle ABC$ . From the similarity we have $\overline{BC}:\overline{DC} = \overline{AC}:\overline{BC}$ , from where we get $\overline{BC}^2 = \overline{DC} \cdot \overline{AC} = 9 \cdot (7+9) = 144$ . Hence $\overline{BC} = 12\text{cm}$ .

From $\triangle BDC \sim \triangle ABC$ we also have $\overline{BD}:\overline{DC} = \overline{AB}:\overline{BC}$ or $\overline{BD}:\overline{AB} = \overline{DC}:\overline{BC} = 9:12 = 3:4$ . Now we obtain $\overline{BD}:\overline{BA} = 3:4$ .

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