Problem

In $△ABC$, when $∠B=2∠C$,

$$AC<2AB.$$

---

$$ $$

$$ $$

$\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$

$$ $$

$$ $$

$$ $$

$$ $$

$$ $$

$$ $$

$$ $$

$$ $$

$$ $$

$$ $$

---

Solution

If we take $E$ on $BC$ so that $∠CAE=∠C$,

$$EA=CE.$$

Since $∠AEB$ is the exterior angle of $△EAC$,

$$∠AEB=∠CAE+∠C,$$

$$∴ \ ∠AEB=2∠C,$$

$$∴ \ ∠AEB=∠B,$$

$$∴ \ AB=EA \ (=CE).$$

Regarding $△EAC$, from problem $0042$,

$$AC<EA+CE,$$

$$∴ \ AC<2AB.$$

$ $

---

Reference

Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2^nd edition), Seikyo-Shinsha, p.14