Given PQRS is a parallelogram, so PQ II SR and PS II QR
Also, AB II PS
To prove OC II SR
Proof In ΔOPS and ΔOAB,
PS II AB [given]
∠POS = ∠AOB [common angle]
∠OSP = ∠OBA [corresponding angles]
\(\therefore \Delta O P S \sim \Delta O A B\) [by AA similariry criterion]
Then, \(\frac{Q R}{A B}=\frac{C R}{C B}\)
\(\Rightarrow \frac{P S}{A B}=\frac{C R}{C B}\) ......(i)
\(\text { In } \Delta C Q R \text { and } \Delta C A B, Q R\|P S\| A B\)
∠QCR = ∠ACB [common angle]
∠CRQ = ∠CBA [corresponding angle]
∴ ΔCQR ∼ ΔCAB [by AA similariry criterion]
Then, \(\frac{Q R}{A B}=\frac{C R}{C B}\) 
\(\Rightarrow \frac{P S}{A B}=\frac{C R}{C B}\) ........(ii)
[since, PQRS is a parallelogram, so PS = QR]
From Eqs. (i) and (ii), we get
\(\frac{O S}{O B}=\frac{C R}{C B} \text { or } \frac{O B}{O S}=\frac{C B}{C R}\)
On subtracting 1 from both sides, we get
\(\frac{O B}{O S}-1=\frac{C B}{C R}-1\)
\(\Rightarrow \frac{O B-O S}{O S}=\frac{C B-C R}{C R} \Rightarrow \frac{B S}{O S}=\frac{B R}{C R}\)
By converse of basic proportionaliry theorem,
SR II OC

\(\text { In } \triangle A P C \text { and } \Delta B P D \text { , }\)
\(\angle A P C=\angle B P D\) [vertically opposite angles]
and \(\angle P A C=\angle P B D\) [alternate angles]
\(\therefore \Delta A P C \sim \Delta B P D\) [by AA similarity criterion]
\(\Rightarrow \frac{A P}{P B}=\frac{A C}{B D}=\frac{P C}{P D}\) ......(i)
[since, corresponding sides of similar triangles are proportional]
\(\text { In } \Delta A P E \text { and } \Delta B P F\)
\(\angle A P E=\angle B P F\) [vertically opposite angles]
and \(\angle P A E=\angle P B F\) [alternate angles]
\(\therefore \Delta A P E \sim \Delta B P F\) [by AA similarity criterion]
\(\Rightarrow \frac{A P}{P B}=\frac{A E}{B F}=\frac{P E}{P F}\) ......(ii)
[since, corresponding sides of similar triangles are proportional]
\(\text { In } \Delta P E C \text { and } \Delta P F D, \angle E P C=\angle F P D\) [vertically opposite angles]
and \(\angle P C E=\angle P D F\) [alternate angles]
\( \therefore \Delta P E C \sim \Delta P F D \) [by AA similarity criterion]
\(\Rightarrow \frac{P E}{P F}=\frac{P C}{P D}=\frac{E C}{F D}\) ........(iii)
[since, corresponding sides of similar triangles are proportional]
From Eqs, (i), (ii) and (iii), we get
\(\frac{A P}{P B}=\frac{A C}{B D}=\frac{A E}{B F}=\frac{P E}{P F}=\frac{E C}{F D}\)
\(\therefore \frac{A E}{B F}=\frac{A C}{B D}=\frac{C E}{F D}\)

In \(\triangle PDQ\), we have
PQ² = PD² + QD²
PD² = PQ² - QD²
= a² - c² .... (i)

In right \(\triangle PDR\), we have
PR² = PD² + DR²
PD² = PR² - DR² = b² - d²
From Eqs.(i) and (ii)
a² - c² = b² - d² \(\Rightarrow \)  a² - b² = c² - d²

Yes, 16 : 1 ; unity of nation, fraternity and patriotism.

Draw the figure according to given condition, then show that both triangles are similar by AAA similarity criterion and then use ratio of sides of both triangles to get the requried length.
10 m