Try this beautiful Rectangle Problem from Geometry, from PRMO 2017.

## Rectangle Problem – Geometry – PRMO 2017, Question 13

In a rectangle $A B C D, E$ is the midpoint of $A B ; F$ is a point on $A C$ such that $B F$ is perpendicular to $A C$; and FE perpendicular to BD. Suppose $\mathrm{BC}=8 \sqrt{3}$. Find AB.

- $9$
- $24$
- $11$

**Key Concepts**

Geometry

Triangle

Trigonometry

## Check the Answer

Answer:$24$

PRMO-2017, Problem 13

Pre College Mathematics

## Try with Hints

We have to find out the value of \(AB\). Join \(BD\). \(BF\) is perpendicular on \(AC\).

Let $\angle \mathrm{BAC}=\theta$

since $\mathrm{E}$ is mid point of hypotenous $\mathrm{AB}$ of right $\Delta \mathrm{AFB}$, therefore $A E=F E=B E$

Can you now finish the problem ……….

Therefore

Therefore$ \angle E F A=\angle F A E=\theta$

and $\angle \mathrm{FEB}=\angle \mathrm{EAF}+\angle \mathrm{EFA}=2 \theta$

$\Rightarrow \angle E B D=90^{\circ}-\angle B E F=90^{\circ}-2 \theta$

But $\angle \mathrm{FAE}=\angle \mathrm{CAB}=\angle \mathrm{DBA}$

Therefore $\theta=90^{\circ}-2 \theta \Rightarrow \theta=30^{\circ}$

Therefore in $\Delta \mathrm{ABC}, \tan \theta=\frac{\mathrm{BC}}{\mathrm{AB}}$

Can you finish the problem……..

Therefore $A B=B C \cot \theta=8 \sqrt{3} \cot 30^{\circ}=24$

## Other useful links

- https://www.cheenta.com/ordered-pairs-prmo-2019-problem-18/
- https://www.youtube.com/watch?v=VLyrlx2DWdA