Question 1012750
{{{drawing(300,220,-1.5,13.5,-1,10,
rectangle(0,0,12,9),
line(0,0,3,9),line(3,9,12,6),
locate(-0.5,0,A),locate(11.8,0,B),
locate(-0.2,9.7,D),locate(2.8,9.7,P),
locate(11.8,9.7,C),locate(12.1,6.5,Q),
locate(1.2,9,x),locate(7,9,12-x),
locate(0.1,5,9),locate(5.5,0,12)
)}}} There are 3 angles at {{{P}}} , whose measures add up to {{{180^o}}} .
{{{DPA+CPB+APQ=180^o}}}
{{{DPA+CPB+90^o=180^o}}}-->{{{DPA+CPB=180^o-90^o}}}-->{{{DPA+CPB=90^o}}}
In right triangle {{{ADP}}} , {{{DPA+DAP=90^o}}} .
{{{system(DPA+CPB=90^o,DPA+DAP=90^o)}}}--->{{{CPB=DAP}}} .
In right triangle {{{PCQ}}} , {{{CQP+CPB=90^o}}}.
{{{system(DPA+CPB=90^o,CBP+CPB=90^o)}}}--->{{{DPA=CQP}}} .
So, triangles {{{ADP}}} and {{{PCQ}}} are similar, because they have congruent pairs of angles.
Since they are similar corresponding sides are proportional:
{{{DP/DA=CQ/CP}}}
{{{x/9=CQ/(12-x)/CQ}}}--->{{{CQ=x(12-x)/9}}}
{{{BQ=BC-CQ}}}--->{{{BQ=9-CQ}}}
{{{system(BQ=9-CQ,CQ=x(12-x)/9)}}}--->{{{highlight(BQ=9-x(12-x)/9)}}}