Question

(iii) AC^2 + AB^2 = 2 AD^2+
+
BC/2
​

Answer 1

Answer:

Let the median from B meet the side AC in D, the median from C meet

AB in E and let P be the point of intersection of the two medians. Recall that the

medians of a triangle meet at a point two-thirds of the way from the vertex to the

midpoint of the opposite side. Thus

Step-by-step explanation:

BP = 2P D and CP = 2P E.

Applying the Pythagorean Theorem in triangles BP E and CP D we have

AB2 = (2BE)

2 = 4·BE2 = 4(BP2+P E2

) = 4(BP2+(CP/2)2

) = 4BP2+CP2

, (1)

AC2 = (2CD)

2 = 4·CD2 = 4(CP2+P D2

) = 4(CP2+(BP/2)2

) = 4CP2+BP2

, (2)

and

BC2 = BP2 + CP2

. (3)

Adding equation (1) and (2) and using (3) gives

AB2 + BC2 = 5(BP2 + CP2

) = 5B