Question

in an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC prove that 9AD^2 =7AB^2.​

Answer 1

Answer

Given that,

∆ABC in which AB = BC = CA

D is the midpoint on BC such that $\sf BD = \dfrac{1}{3} BC$

To prove,

9AD² = 7AB²

Construction,

Draw AL⊥BC

Proof,

In right angled triangles ALB and ALC we have,

$\sf AB = AC \: \: (given)$

$\sf AL = AL \: \: (common)$

By RHS axiom,

$\sf \triangle ALB \cong \triangle ALC$

So, $\sf BL = CL$

Thus,

$\sf \dashrightarrow BD = \dfrac{1}{3} BC \: \: and \: \: BL = \dfrac{1}{2} BC$

By Pythagoras theorem,

In ∆ALB, ∠L = 90°

So, AB² = AL² + BL²

In ∆ALD, ∠L = 90°

So, AD² = AL² + DL²

$\sf \dashrightarrow {AL}^{2} + (BL - BD)^{2}$

$\sf \dashrightarrow {AL}^{2} + {BL}^{2} + {BD}^{2} - 2BL . BD$

$\sf \dashrightarrow ({AL}^{2} + {BL}^{2}) + {BD}^{2} - 2BL . BD$

$\sf \dashrightarrow {AB}^{2} + {BD}^{2} - 2BL . BD$

$\sf \dashrightarrow {BC}^{2} + \bigg( \dfrac{1}{3} BC \bigg)^{2} - 2 \bigg( \dfrac{1}{2} BC \bigg) . \dfrac{1}{3} BC$

$\sf \dashrightarrow {BC}^{2} + \dfrac{1}{9} {BC}^{2} - \dfrac{1}{3} {BC}^{2}$

BC = AB. So,

$\sf \dfrac{7}{9} {BC}^{2} = \dfrac{7}{9} {AB}^{2}$

Thus, 9AD² = 7AB².

Answer 2

Answer:

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