Trigonometric rectangle

Question

In the rectangle ABCD, $$1. \, BE = EF = FC = AB$$ $$2. \, \angle AEB = \beta , \angle AFB = \alpha , \angle ACB = \theta. $$ Prove that $\alpha + \theta = \beta$.

I have so far obtained that - $$1. \cos\beta = \sin \beta$$ $$2.\cos\alpha = 2\sin\alpha$$ $$3. \cos \theta = 3\sin \theta$$ But I am not able to understand what to do next. Please help.

Answer 1

It is clear that the three angles are acute. Moreover,

$$\tan\beta=2\tan\alpha=3\tan\theta$$

$$\begin{align} \tan(\alpha+\theta)&=\frac{\tan\alpha+\tan\theta}{1-\tan\alpha\tan\theta}\\ &=\frac{\frac56\tan\beta}{1-\frac{\tan^2\beta}6}\\ &=\frac{5\tan\beta}{6-\tan^2\beta} \end{align}$$

So the statement is true only if $6-\tan^2\beta=5$, that is, only if $\tan\beta=1$ or $\beta=45^\circ$.

Answer 2

You have $\beta= 45degree$ because an isoceles right triangle ABE. then $\tan(\alpha + \theta)$=$(\tan\alpha+\tan\theta)$,$1-\tan\alpha\tan\theta =$(5/6 $\tan\beta$)$1-\tan^2\beta$=$5\tan\beta$ $6-\tan^2\beta$ gives $\tan(\alpha + \theta)=1=\tan\beta$ $\alpha + \theta = \beta$