$ABCD$ is a quadrilateral with right angled at $A$ and $C$. Points $E$ and $F$ are on the diagonal $AC$ such that $DE$ and $BF$ are both perpendicular to $AC$. They are asking $BF$. If $AE=3$, $DE=5$, $CE=7$.
This is a sum based on the principles of similarity and basic proportionality theorem and basic geometry. I tried but I couldn't get it. I tried similarity between ∆DEF and ∆BFC but I couldn't find FC so I was unable to solve it
On
There is no need for trigonometry at all. Using Doubtnut's diagram, $\angle FAB = 90º - \theta$, and since $\angle AFB$ is right as well, $\angle ABF = \theta$. Therefore $\Delta ADE \sim \Delta BAF$.
Using the same reasoning, $\Delta DCE \sim \Delta FBC$.
Now use the similarity of the triangles to construct two equations. At first, there might be too many unknown sides, but you can, in fact, express all the remaining sides in terms of $AF$. Solve for $AF$ and use this to find $BF$.
Can you continue?
The full solution is hidden behind the spoiler:
$$\Delta ADE \sim \Delta BAF \Rightarrow \frac{3}{5} = \frac{BF}{AF} \tag{1}$$ $$\Delta DCE \sim \Delta FBC \Rightarrow \frac{5}{7} = \frac{CF}{BF} \tag{2}$$ From $(1)$, $BF = \frac{3}{5} AF$. Since $AC = AE + CE = 10$, $CF = 10 - AF$. Combining these two facts together: $$\frac{5}{7} = \frac{10-AF}{(3/5) AF} \Rightarrow 3AF = 70 - 7AF \Rightarrow AF = 7$$ and since $BF = \frac{3}{5} AF$, $BF = \frac{21}{5}$.
This is the figure of the given situation. Let $EF=x$.
In $\triangle DAE$, $$\tan \theta=\frac{DE}{AE}=\frac{5}{3}$$ In $\triangle BFA$, $\angle FAB=(90°-\theta)$ $$\therefore \tan(90°-\theta)=\frac{BF}{AF}=\frac{BF}{3+x}\dots\tag{1}$$
Now, in $\triangle DCE$, $$\tan\alpha=\frac{5}{7}$$ In $\triangle CFB$, $\angle FCB=(90°-\alpha)$ $$\therefore \tan(90°-\alpha) =\frac{BF}{FC}=\frac{BF}{7-x}\dots\tag{2}$$
Divide equation $1$ and $2$, which gives us $x=4$. Now, you can find $BF$.