1. Since $\triangle PQR$ is right-angled at $R$, we have $\angle R=\frac{\pi}{2} \implies P+Q=\frac{\pi}{2}.$

2. Let $\alpha=\tan\frac{P}{2}, \qquad \beta=\tan\frac{Q}{2}.$ These are given as the roots of $ax^2+bx+c=0,\qquad a\ne 0.$

So by Vieta's formulas, $\alpha+\beta=-\frac{b}{a}, \qquad \alpha\beta=\frac{c}{a}.$

3. Use the angle relation: $\frac{P}{2}+\frac{Q}{2}=\frac{P+Q}{2}=\frac{\pi}{4}.$ Therefore, $\tan\left(\frac{P}{2}+\frac{Q}{2}\right)=\tan\frac{\pi}{4}=1.$

Now, \tan\left(\frac{P}{2}+\frac{Q}{2}\right)=\frac{\tan\frac{P}{2}+\tan\frac{Q}{2}}{1-\tan\frac{P}{2}\tan\frac{Q}{2}}=rac{\alpha+\beta}{1-\alpha\beta}. Thus, $\frac{\alpha+\beta}{1-\alpha\beta}=1.$

4. Simplify: $\alpha+\beta=1-\alpha\beta.$ So, $\alpha+\beta+\alpha\beta=1.$

5. Substitute Vieta's values: $-\frac{b}{a}+\frac{c}{a}=1.$ Multiply by $a$: $-b+c=a.$ Hence, $c=a+b.$

6. Check options:

• A: $a=b+c$ ❌
• B: $c=a+b$ ✅
• C: $b=c$ ❌
• D: $b=a+c$ ❌

Therefore, the correct option is B.