Question

Please solve the question ​

Answer 1

Answer:

by theorem of class 9 that states if two triangles are between same parallels and on same base so hence proved that area of abc = area of abd

Answer 2

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here , a ² + b² + c² – ab – bc – ca = 0

we have to prove a = b= c

now, the solution .....

by , multiplying both sides with 2, we get the result.

2( a ²+ b² + c² – ab – bc – ca) = 0

⇒ 2a² + 2b²+ 2c² – 2ab – 2bc – 2ca = 0

⇒ (a² – 2ab + b²) + (b² – 2bc + c²) + (c² – 2ca + a²) = 0

⇒ (a –b)² + (b – c)² + (c – a)² = 0

hence, we can say ,(a - b)² = (b - c)² = (c - a)² = 0

therefore,

we can say, (a - b)² = 0 ---------- (1)

(b - c)² = 0 ---------- (2)

(c - a)² = 0 ---------- (3)

therefore, by Simplifying Equ. (1), we get

(a - b)² = 0

now Taking the Square Root on both sides, we have

a - b = 0

a = b ---------- (4)

similarly, Simplifying Equ. (2), we get

(b - c)² = 0

Taking Square Root on both sides, we have

b - c = 0

b = c ---------- (5)

again, Simplifying Equ. (3), we have

(c - a)² = 0

Taking Square Root on both sides, we have

c - a = 0

c = a ---------- (6)