In an equilateral triangle PQR , if PT is perpendicular to QR then prove that 3PQ² = 4PT^2.

Answer :

Given  PQR is a equilateral triangle 

And PT is perpendicular to QR , and we know " Every altitude is also a median and a bisector. " 

So,

PQ  =  PR   = QR  
And
QT  =  TR  = $\frac{QR}{2}$

Now we apply Pythagoras theorem In $∆$ PQT , and get

PQ^​2  = PT²  + QT²

$\Rightarrow$PQ^​2 = PT² + ( $\frac{QR}{2}$ )²                              ( As we know  QT   = $\frac{QR}{2}$ )

$\Rightarrow$PQ^​2 = PT² +  $\frac{Q{R}^2}{4}$
taking L.C.M. we get 

$\Rightarrow$PQ²  = $\frac{4 P{T}^2 + Q{R}^2}{4}$

$\Rightarrow$4 PQ² = 4PT²  + QR² 

$\Rightarrow$4 PQ² -  QR² = 4PT² 

$\Rightarrow$4 PQ² -  PQ² = 4PT²                                            ( As we know PQ  =  QR )

$\Rightarrow$3 PQ² = 4PT²                                                      ( Hence proved )