\(ABCD\) একটি আয়তক্ষেত্র। \(AD\) বাহুর মধ্যবিন্দু \(E\) এবং \(ED\)-এর মধ্যবিন্দু \(F\)। \(AB\) বাহুকে \(CE\) রেখা \(G\) বিন্দুতে এবং \(CD\) বাহুকে \(BF\) রেখা \(H\) বিন্দুতে ছেদ করে। \(\triangle BCG\) এবং \(\triangle BCH\)-এর ক্ষেত্রফলের অনুপাতকে যদি লঘিষ্ঠ আকারে \(\frac{m}{n}\) হিসেবে লেখা যায়, তবে \(10m+10n+mn\)-এর মান বের কর।
Consider rectangle \(ABCD\). Let \(E\) be the midpoint of side \(AD\) and let \(F\) be the midpoint of \(ED\). Let \(G\) be the intersection of \(CE\) with the line \(AB\) and let \(H\) be the intersection of \(BF\) with line \(CD\). The ratio of areas of the \(\triangle BCG\) and \(\triangle BCH\) can be expressed as \(\frac{m}{n}\) in lowest terms. Compute \(10m+10n+mn\).
BdMO National Junior 2020 P2
This section is intentionally left blank.
Re: BdMO National Junior 2020 P2
$\triangle AGE$ is similar to $\triangle GBC$. Since, $AE = \frac{1}{2} AD = \frac{1}{2} BC$, $AG = \frac{1}{2}BG $ or $AG = AB$.
Again, $\triangle FHD$ is similar to $\triangle HBC$ and $FD = \frac{1}{4}AD $. So we have $DH = \frac{1}{3}CD$
Now, $BG = AB+AG = 2AB$ whereas $CH = CD+DH= \frac{4}{3}CD = \frac{4}{3}AB$. $\triangle BCG $and$ \triangle BCH $ share same base $BC$, so ratio of their area $$\frac{m}{n}=\frac{2AB}{\frac{4AB}{3}} = \frac{3}{2}$$
Again, $\triangle FHD$ is similar to $\triangle HBC$ and $FD = \frac{1}{4}AD $. So we have $DH = \frac{1}{3}CD$
Now, $BG = AB+AG = 2AB$ whereas $CH = CD+DH= \frac{4}{3}CD = \frac{4}{3}AB$. $\triangle BCG $and$ \triangle BCH $ share same base $BC$, so ratio of their area $$\frac{m}{n}=\frac{2AB}{\frac{4AB}{3}} = \frac{3}{2}$$