Question

y shouldn't we plot AD on RHS
proper answer is needed

Answer 1

Solution :

This Question has taken from the triangle chapter of class 10th CBSE . Well , your question is a general question . Let us clarify the doubt along with prove of 9AD^2 = 7AB^2  where BD = 1/3 BC .

Question :

In  a equilateral triangle ABC , D is a point on side BC such that BD = 1/3 BC . Prove that 9AD^2 = 7AB^2 ?

Answer :

At the very initial stage we draw a figure of it . Make a equilateral triangle ABC . Now , according to the question" D is a point on BC " It means that a perpendicular can be drawn from A to D and from AD we construct the point E .  We can also suppose AE is a perpendicular where A can be extended to D . This is a basic that we can't consider AD on R.H.S according to our earlier classes concepts . We say ABCDE not ABCED . According to the data provided to us that BD = 1/3 BC . After take a Glance at the figure let us begin the proof .

Given :

BD = 1/3 BC

Δ ABC is an equilateral triangle

To Prove :

9AD^2 = 7AB^2

Construction :

Draw AE ⊥ BC

Proof :

In Δ ADE

⇒ AD^2 = AE^2 + DE^2 --> 1

In Δ ABE

⇒ AB^2 = AE^2 + BE^2 --> 2

From equation (1) and (2)

⇒ $AD^2 = AB^2 + BE^2 + DE^2$

⇒ $AD^2 = AB^2 - (\frac{BC}{2})^2 + ( BE - BD)^2$

⇒ $AD^2 = AB^2 - \frac{BC^2}{4} + BE^2 + BD^2 - 2BE . BD$

⇒ $AD^2 = AB^2 - \frac{BC^2}{4} + (\frac{BC}{2} )^2 + (\frac{1}{3} BC)^2 - 2 . \frac{1}{3} BC . \frac{BC}{2}$

⇒ $AD^2 = AB^2 - \frac{BC^2}{4} +\frac{BC^2}{4} + \frac{1}{9} BC^2 - \frac{BC^2}{3}$

⇒ $AD^2 = AB^2 - \frac{2BC^2}{9}$

⇒ $AD^2 = AB^2 - \frac{2AB^2}{9}$

⇒ $9AD^2 = 9AB^2 - 2AB^2$

⇒ $9AD^2 = 7AB^2$

Answer 2

Step-by-step explanation:

At the very initial stage we draw a figure of it . Make a equilateral triangle ABC . Now , according to the question" D is a point on BC " It means that a perpendicular can be drawn from A to D and from AD we construct the point E . We can also suppose AE is a perpendicular where A can be extended to D . This is a basic that we can't consider AD on R.H.S according to our earlier classes concepts . We say ABCDE not ABCED . According to the data provided to us that BD = 1/3 BC . After take a Glance at the figure let us begin the proof .

Given :

BD = 1/3 BC

Δ ABC is an equilateral triangle

To Prove :

9AD^2 = 7AB^2

Construction :

Draw AE ⊥ BC

Proof :

In Δ ADE

⇒ AD^2 = AE^2 + DE^2 --> 1

In Δ ABE

⇒ AB^2 = AE^2 + BE^2 --> 2