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In an equilat...

Question

In an equilateral triangle $A B C$ , the side $B C$ is trisected at $D$ i.e., $B D = \frac{1}{3} B C$ . Prove that $9 A D^{2} = 7 A B^{2}$ .

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Solution

$G i v e n : △ A B C i s a n e q u i l a t e r a l t r i a n g l e, w h o s e s i d e B C i s t r i s e c t e d a t D . T o P r o v e : 9 {A D}^{2} = {7 A B}^{2} P r o o f : △ A B C b e a n e q u i l a t e r a l t r i a n g l e a n d D b e t h e p o i n t o n B C s u c h t h a t B D = \frac{1}{3} B C (G i v e n) D r a w A E ⊥ B C, J o i n A D . B E = E C (A l t i t u d e d r a w n f r o m a n y v e r t e x o f a n e q u i l a t e r a l t r i a n g l e b i s e c t s t h e o p p o s i t e s i d e) S o, B E = E C = \frac{B C}{2} I n △ A B C {A B}^{2} = {A E}^{2} + {E B}^{2} . . . (1) {A D}^{2} = {A E}^{2} + {E D}^{2} . . . (2) F r o m (1) a n d (2) {A B}^{2} = {A D}^{2} - {E D}^{2} + {E B}^{2} \Rightarrow {A B}^{2} = {A D}^{2} \frac{{B C}^{2}}{36} + \frac{{B C}^{2}}{4} (∵ B D + D E = {B C}^{2} \Rightarrow \frac{B C}{3} + D E = \frac{B C}{2} \Rightarrow D E = \frac{B C}{6}) \Rightarrow {A B}^{2} + \frac{{B C}^{2}}{36} - \frac{{B C}^{2}}{4} = {A D}^{2} (∴ E B = {B C}^{2}) \Rightarrow {A B}^{2} + \frac{{A B}^{2}}{36} \frac{{A B}^{2}}{4} = {A D}^{2} (∴ A B = B C) \Rightarrow \frac{36 {A B}^{2} + {A B}^{2} - 9 {A B}^{2}}{36} = {A D}^{2} \Rightarrow \frac{28 {A B}^{2}}{36} = {A D}^{2} \Rightarrow 7 {A B}^{2} = 9 {A D}^{2} H e n c e P r o v e d$

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In an equilateral triangle ABC, D is a point on side BC such that BD = BC. Prove that 9 AD² = 7 AB².

Q. Question 15
In an equilateral triangle ABC, D is a point on side BC such that

B D = \frac{1}{3} B C

9 A D^{2} = 7 A B^{2}

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