Question

1. In Fig. 10.10, ABC is a triangle
with D as the mid point of BC.
Name AD and AE. IS AD = AE?​

Answer 1

Answer:

Step-by-step explanation:

AEB is a right angled triangle.

Therefore, by Pythagoras theorem,

AB 2= AE2 + BE2 --------------- (1)

Now, AED is a right angled triangle.

Therefore,

=> AE2 + ED2= AD2  (Pythagoras theorem)

=> AE2 = AD2 - ED2 -------------- (2)

Now, BE = BD - ED ---------------- (3)

Substituting (2) and (3) in (1), we get,

AB2= AD2 - ED2 + (BD - ED)2

 =AD2 – ED2 + BD2 -2BD*ED + ED2  

 = AD2 – ED2 + BD2 -2BD*ED + ED2  

Now, BD = BC/2  (since, D is midpoint)

Therefore,

AB2= AD2 + BD2 -2BD*ED

 =AD2 + (BC/2)2 – 2(BC/2)*ED

 = AD2 + (BC/2)2 – 2(BC/2)*ED

 = AD2 – BC*DE + ¼ BC2