Question

The perpendicular from A on side DC of a $△ABC$. Intersect BC
at D such that $DB=3\times CD$. Prove that $2{AB}^2=2{AC}^2+{BC}^2$

Answer 1

Let $\angle ACD=\theta \angle ABC=\alpha$ then ,$h=x\tan \theta =3\tan \alpha ⟹\tan \theta =\tan \alpha$

$\frac{\sin \theta }{\cos \theta }=\frac{3\sin \alpha }{\cos \alpha }⟹\frac{AB}{\frac{A{C}^2+B{C}^2-A{B}^2}{2.AC.BC}}=\frac{3AC}{\frac{A{B}^2+B{C}^2-A{C}^2}{2.AB.AC}}⟹2A{B}^2=2A{C}^2+B{C}^2$