Question

Help me in solving 6 part plz help by congurency

Answer 1

Construct a triangle ABC and let AD be the bisector of angle BAC.

Since angle BAC is bisected

=> angle BAD = angle DAC

Now, in ∆ABD and ∆ACD,

AB = AC (Given)

angle BAD = angle DAC (Given)

AD = AD (Common)

=> ∆ ABD is congruent to ∆ ACD. (SAS)


Since ∆ABD and ∆ACD are congruent,

=> BD = DC (c.p.c.t)

=> AD bisects BC .........(1)

Again,

=> angle ADB = angle ADC (c.p.c.t)

and,

angle ADB + angle ADC = 180° (linear pair)

but angle ADB = angle ADC

=> angle ADB + angle ADB = 180°

=> 2 × angle ADB = 180°

=> angle ADB = 180/2

=> angle ADB = 90°

And angle ADB = angle ADC

=> angle ADB = angle ADC = 90°

Hence, AD is also perpendicular to BC.... (2)


From (1) and (2), we can say that,

AD, that is, the bisector of angle BAC is also the perpendicular bisector of BC (base)

Hence Proved :)