• # question_answer Passage: (Q. - 73) A triangle ABC is inscribed in a circle with AL, BM & CN as the diameters of circumcircle, then If AL, BM & CN are angular bisectors of triangle, then if ${{\Delta }_{1}}=$area of $\Delta ABC\And {{\Delta }_{2}}=Ar.$of $\Delta LMN,$then A) ${{\Delta }_{1}}={{\Delta }_{2}}$                             B) ${{\Delta }_{1}}\ge {{\Delta }_{2}}$ C) ${{\Delta }_{1}}\le {{\Delta }_{2}}$                         D) ${{\Delta }_{1}}>{{\Delta }_{2}}$

From figure $\angle ABL=\angle ACL={{90}^{0}}$$\angle ALC=\angle ABC=\angle B$(angle in same segment) Similarly, $\angle ALB=\angle ACB=\angle C$ In $\angle ABL,BL=c\cot \,C\And LC=b\,cot\,B$ $\therefore$Area of $\Delta BLC=\frac{BL.CL.BC}{4R}=\frac{abc}{4R}\cot B.\cot C=2{{r}^{2}}\sin A\cos B\cos C$Hence, sum of area of $\Delta 'sBLC,CMA\And ANB$ is $=2{{R}^{2}}[sinAcosBcosC+sinBcosAcosC$$+\sin C\cos A\cos B]$ $=2{{R}^{2}}\sin A\sin B\sin C=\frac{abc}{4R}={{\Delta }_{1}}$ Also,$\angle ALN=\angle ACN=\frac{C}{2}$ $\angle ALM=\angle ABM=\frac{B}{2}$ $\therefore$$\angle NLM=\frac{B+C}{2}={{90}^{0}}-\frac{A}{2}$ $\therefore$${{\Delta }_{2}}(Ar.of\,\Delta LMC)$ $=2{{R}^{2}}\sin \left( \frac{\pi }{2}-\frac{A}{2} \right)\sin \left( \frac{\pi }{2}-\frac{B}{2} \right)\sin \left( \frac{\pi }{2}-\frac{C}{2} \right)$ $=2{{R}^{2}}\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$ $\therefore$${{\Delta }_{1}}\le {{\Delta }_{2}}$