Question

prove that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides using the above result Prove the following in triangle abc xy is parallel to BC and it divides triangle ABC into two equal parts prove that BX by a b is equal to root 2 minus 1 by root 2

Answer 1

Answer:

Step-by-step explanation:

Sides of triangle A are a , b , c

Let say k is ratio

so sides of new triangles are  ak , bk  , ck

Area of 1st triangle = √(s(s-a)(s-b)(s-c))

s = (a+b+c)/2

S2 = (ak +  bk + ck)/2 = k(a+b+c)/2 = ks

Area of 2nd Triangle = √(ks(ks-ak)(ks-bk)(ks-ck))

= √(ksk(s-a)k(s-b)k(s-c))

=√(k^4)(s(s-a)(s-b)(s-c))

= k^2 √(s(s-a)(s-b)(s-c))

= k^2 (area of 1st Triangle)

k was ratio of side  

and k^2 is ration of Area

QED

ABC is similar to Axy

Area of axy = (1/2 ) Area of ABC

using above proof

so Ax / AB =  1/√2

Ax = AB/√2

Bx/AB = (AB-Ax)/AB

= (AB - AB/√2)/AB

= 1 - 1/√2

= (√2-1)/√2

QED