Library / Ukrainian National Mathematical Olympiad /13 of 201

Ukrainian National Mathematical Olympiad

Geometry Difficulty 4.9 AIME Prove it Ukraine

For a quadrilateral $ABCD$ , $\angle ABD = \angle DBC$ and $AD = CD$ . Let $DH$ be the height of $\triangle ABD$ . Prove that $|BC - BH| = HA$ .

(Danylo Khilko)

Solution

On the ray $BA$ , we put down a segment $BE = BC$ . If point $E$ belongs to $AB$ (fig. 19), then $\triangle BCD = \triangle BED$ due to two pairs of equal sides and the angle between them. Then, $AD = CD = ED$ , which yields that $\triangle ADE$ is isosceles. There, $HD$ is the height and, hence, the median. Therefore,

$AH = HE = HB - BE = HB - BC \Rightarrow BC = BH - HA.$

If point $A$ belongs to the segment $BE$ (fig. 20), then, analogously, $\triangle BCD = \triangle BED$ and

$AH = HE = BE - HB = BC - HB \Rightarrow BC = BH + HA.$

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.