9AD^2=7AB^2

In an equilateral ABC, D is a point on side BC such that BD equals 1 third BC. Prove that 9AD2 = 7AB2. Given : An equilateral triangle ABC such that To Prove : 9AD2 = 7AB Given : An equilateral triangle ABC such that BD equals 1 third BC.To Prove : 9AD2 = 7AB2 Const: Draw AE ⊥ BC. Proof: In right triangles ABE and ACE, we have AE = AE [common] ∠AEB = ∠AEC [90°] and AB = AC [∆ABC is an equilateral] Therefore, by using RHS congruent condition space space increment ABE approximately equal to increment ACE rightwards double arrow BE = CE [by CPCT] rightwards double arrow BE equals CE equals 1 half BC In right triangle ADE, we have space AD squared space equals space AE squared plus DE squared rightwards double arrow AD squared space equals space AE squared plus left parenthesis BE minus BD right parenthesis squared rightwards double arrow space space space AD squared equals AE squared plus BE squared plus BD squared minus 2 BD. BE rightwards double arrow AD squared space equals space left parenthesis AE squared plus BE squared right parenthesis plus BD squared minus 2 BD. BE open square brackets because space space BD equals 1 third BC space and space BE space equals space CE space equals space 1 half BC close square brackets rightwards double arrow space space space AD squared space equals space AB squared plus open parentheses 1 third BC close parentheses squared minus 2 cross times 1 third BC cross times 1 half BC rightwards double arrow space space space AD squared space equals space AB squared plus 1 over 9 BC squared minus BC squared over 3 rightwards double arrow AD squared equals fraction numerator 9 AB squared plus BC squared minus 3 BC squared over denominator 9 end fraction rightwards double arrow 9 AD squared space equals space 9 AB squared minus 2 BC squared But AB = BC = CA rightwards double arrow <pre>uncaught exception: mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied

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

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Given : An equilateral triangle ABC such that To Prove : 9AD² = 7AB²

Const: Draw AE ⊥ BC.
Proof: In right triangles ABE and ACE, we have
AE = AE    [common]
∠AEB = ∠AEC    [90°]
and    AB = AC
[∆ABC is an equilateral]
Therefore, by using RHS congruent condition
                                       
                                          BE = CE         [by CPCT]
                                          
In right triangle ADE, we have
                          
                     
                  
                     
                            


  
             
But             AB = BC = CA