question_answer

ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Then AC + AD = ___.

A)  BC                              

B)         2BC                

C)         3BC                            

D)         None of these

Correct Answer: A

Solution :

Given a right angled\[\Delta ABC,\]where AB = AC and CD is the bisector of \[\angle C.\] Let us draw\[DE\bot BC.\] In right angled\[\Delta ABC,AB=AC,\]and \[\angle A={{90}^{o}}\] Now, in\[\Delta DAC\]and\[\Delta DEC,\] \[\angle A=\angle 3\]                   [Each\[{{90}^{o}}\]] \[\angle 2=\angle 1\]       [\[\because \]CD is the bisector of\[\angle C\]] DC= DC           [Common] \[\therefore \]\[\Delta DAC\cong \Delta DEC\][By AAS congruency] \[\Rightarrow \]\[DA=DE\]                                                 ?(1) and AC = EC                [By C.P.C.T]      ?(2) Also, in \[\Delta ABC,AB=AC\Rightarrow \angle C=\angle B\]      ?(3) [Angles opposite to equal sides are equal] Again, in \[\Delta \Alpha \Beta C,\angle A+\angle B+\angle C={{180}^{o}}\] [by angle sum property of triangle] \[\Rightarrow \]\[{{90}^{o}}+\angle B+\angle B={{180}^{o}}\] [From (3)] \[\Rightarrow \]\[2\angle B={{90}^{o}}\Rightarrow \angle B={{45}^{o}}\] In \[\Delta BED,\angle 5={{180}^{o}}-(\angle B+\angle 4)\] \[\therefore \] \[\angle 5={{180}^{o}}-({{45}^{o}}+{{90}^{o}})={{180}^{o}}-{{135}^{o}}={{45}^{o}}\] \[\angle B=\angle 5\Rightarrow DE=BE\]                           ?(4) [Sides opposite to equal angles are equal from (1) and (4)]. DA = DE = BE                                      ?(5) \[\because \] BC = EC + BE = AC + AD [from (2) and (5)] Thus, AD + AC = BC