Question

In AABC, ZBAC = 90°, seg BL and seg CM
are medians of AABC. Then prove that
4(BL^2+ CM^2) - 5BC^2​

Answer 1

Step-by-step explanation:

Given: In triangle ABC,angle BAC =90°and BL and CM are medians of triangle ABC

To Prove:4(BL^2+ CM^2) - 5BC^2

Proof:BL is the median

AL=CL=AC/2 ...(i)

CM is the median

AM =MB=AB/2 ...(ii)

In right triangle BAC

BC^2= AB^2+AC^2 (iii)

In right triangle BAL

BL^2= AB^2+AL^2

BL^2= AB^2+(AC/2)^2 ... [From(i)]

BL^2= AB^2+AC^2/4

4BL^2= 4AB^2+AC^2 ..(iv)

In right triangle MAC

CM^2=AM^2+AC^2

CM^2= (AB/2)^2+AC^2 [From(ii)]

CM^2= AB^2/4+AC^2

4CM^2= AB^2+4AC^2 ...(v)

Adding (iv) and (v)

4BL^2+4CM^2= 4AB^2+AC^2+AB^2+4AC^2

4(BL^2+CM^2) =5AB^2+5AC^2

4(BL^2+CM^2) =5(AB^2+AC^2)

4(BL^2+CM^2) =5BC^2 [From(iii)]

4(BL^2+CM^2) -5BC^2

Hence proved

Answer 2

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Hope it helps!

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