Question 1203590
Let {{{ ABCD }}}be a trapezium, 
{{{ E }}}the midpoint of leg {{{ AD }}}
{{{ F }}}the midpoint of leg {{{ BC}}}

Prove that{{{  EF=(1/2)(AB+DC)}}}

<a href="https://ibb.co/wKp3ZdT"><img src="https://i.ibb.co/wKp3ZdT/Capture.jpg" alt="Capture" border="0"></a>



{{{ ABCD}}} is trapezium in which {{{ AB}}}||{{{ DC}}}.

{{{ EF}}} is parallel to side {{{ DC}}}. 

Then we have {{{ AB}}}||{{{ DC}}}||{{{ EF}}}.

Hence we have also trapezium {{{ ABFE}}} and trapezium {{{ EFCD}}}.

Let {{{ AP}}} be the perpendicular to {{{ DC}}} and this intersects {{{ EF}}} at {{{ Q}}}. 
{{{ AQ}}} will be perpendicular to {{{ EF}}}.

For △{{{ APD}}} and △{{{ AQE }}}we have  

{{{ AD/EA= AP/AQ=2}}} 

This gives {{{ AP=2AQ }}}  
i.e, {{{ AQ=QP}}} 


Consider the area  we have area {{{ ABCD}}} = area {{{ ABFE}}} + area {{{ EFCD }}}  

                  
{{{ ( 1/2)AP*(AB+DC)=( 1/2)AQ*(AB+EF)+( 1/2)QP*(EF+DC)}}} 

=> {{{ AP(AB+DC)=AP*(AB/2)+AP*(EF/2)+AP*(EF/2)+AP*(DC/2)}}} 
​
         
=> {{{ AP*(AB/2)+AP*( DC/2)}}} 
​

 ={{{ AP*(EF/2) +AP* (EF/2)}}} 
​

={{{ AP*((AB+DC)/2)}}} 
​
 ={{{ AP*EF}}} 

=>

{{{ EF= (AB+DC)/2}}} 
​