Question

In the given figure, ABC is a triangle right-angled at B such that ∠BCA = 2 ∠BAC. Show that AC = 2BC.

Answer 1

Produce CB to D such that BD = BC.
Join AD.
$$\begin{gathered} \mathrm{In}△\mathrm{ABC}\mathrm{and}△\mathrm{ABD},\mathrm{we}\mathrm{have}: \\ \mathrm{BC}=\mathrm{BD}(\mathrm{By}\mathrm{construction}) \\ \mathrm{AB}=\mathrm{AB}\left(\mathrm{Common}\right) \\ \angle \mathrm{ABC}=\angle \mathrm{ABD}=90° \\ \therefore △\mathrm{ABC}\cong △\mathrm{ABD}(\mathrm{SAS}\mathrm{criterion}) \\ \Rightarrow \angle \mathrm{CAB}=\angle \mathrm{DAB}\left(\mathrm{CPCT}\right) \\ \Rightarrow \mathrm{AC}=\mathrm{AD}\left(\mathrm{CPCT}\right) \\ \therefore \angle \mathrm{CAD}=\angle \mathrm{CAB}+\angle \mathrm{BAD} \\ =\mathrm{x}°+\mathrm{x}°=2\mathrm{x}° \end{gathered}$$
But $AC=AD$
$\mathrm{i}.\mathrm{e}.,\angle ACD=\angle ADB=2x°(\because \angle BCA=2\angle BAC)$
Therefore, triangle ACD is an equilateral triangle.
$$\begin{gathered} \mathrm{Then}AC=CD \\ \Rightarrow AC=2BC \end{gathered}$$