Question

D and E are midpoints of the sides BC and CA respectively of a triangle ABC right angled at C prove that: 4AD^2 = 4AC^2 + BC^2​

Answer 1

Step-by-step explanation:

from the given figure,

In ∆ABC, D and E are midpoints of the sides BC and CA respectively of this triangle and ∆ABC is right angled at C

so from given condition ,

CD =1/2BC and CE =1/2AC

and CD=BC , CE=AD

Let's in ∆ABC, there is also a ∆ADC

so from Pythagoras theorem,

AD^2 = AC^2 + CD^2

AD^2=(2CE)^2 + (1/2BC )^2

AD^2=4CE^2+1/4BC^2

4AD^2=4CE^2+BC^2

but we know that

CE=AC

so, 4AD^2=4 AC^2+BC^2