Find angle of a rhombus

Question

In a rhombus $ABCD$ let $O$ be the intersection of diagonals. Given that $AB + BO = AC$, find $\sin(BAD)$

Answer 1

Note: $$\begin{cases}AO^2+BO^2=AB^2 \\ AB+BO=2AO \end{cases} \Rightarrow \begin{cases} AO=\frac{4AB}{5} \\ BO=\frac{3AB}{5}\end{cases}.$$ Hence: $$\sin{(BAD)}=\sin{(2BAO)}=2\sin{(BAO)}\cos{(BAO)}=2\cdot \frac{BO}{AB}\cdot \frac{AO}{AB}=2\cdot \frac35 \cdot \frac45=\frac{24}{25}.$$

Answer 2

Hint:

If $\angle BAD=4x,0<4x<180^\circ\iff0<x<?$

$$AC=2AB\cos2x,BO=AB\sin2x$$

$$\implies2\cos2x=\sin2x+1$$

Use Weierstrass Substitution

Answer 3

Let $AB=x$ and $BO=y$.

THus, by the given we obtain: $$x+y=2\sqrt{x^2-y^2}$$ or $$3x^2-2xy-5y^2=0$$ or $$3x-5y=0,$$ which gives $$\sin\measuredangle BAC=\frac{y}{x}=\frac{3}{5}.$$ Id est, $$\sin\measuredangle BAD=2\sin\measuredangle BAC\cos\measuredangle BAC=2\cdot\frac{3}{5}\cdot\frac{4}{5}=\frac{24}{25}.$$