Question 1129840

{{{drawing ( 600, 600, -10, 10, -10, 10,

line(1,0,7,0), line(1,0,-1,5),line(-1,5,7,0), line(4,10,4,-10),
locate(-.5,2.8,M),locate(-1.5,5.2,A),locate(1,-0.2,C), locate(7,-.2,B),
locate(1.5,3.8,P),locate(4.2,2.2,Q),locate(4.2,5.2,E),locate(4.2,-0.2,N),
line(1,0,4,2),line(1,0,1.5,3.5),locate(1.6,2.8,78),locate(3,2,62),
graph( 600, 600, -10, 10, -10, 10, (2/5)x+9/3.333 )) }}}




Let {{{M}}} be the midpoint of side {{{AC}}}, then by SAS similarity postulate, Δ {{{AMP }}}is congruent to Δ {{{CMD}}}.

Thus, {{{m}}} < {{{APM}}} = {{{m}}} < {{{CPM}}}

But {{{m}}} < {{{APM }}}+ {{{m}}} < {{{CPM}}} + {{{m}}} < {{{CPQ}}} = {{{180}}}°

&#8658; {{{2m}}} < {{{CPM }}}= {{{180 - 78 = 102}}}°

&#8658; {{{m}}} < {{{CPM = 51}}}°

But {{{m}}} < {{{ACP }}}= {{{90}}}° - {{{m}}} < {{{CPM}}} = {{{90 - 51 = 39}}}°.

Similarly, let {{{N}}} be the midpoint of side {{{BC}}}, then by {{{SAS}}} similarity postulate, &#916; {{{CNQ}}} is congruent to &#916; {{{BNQ}}}.

Thus, {{{m}}} < {{{CQN}}} ={{{ m}}} < {{{BQN}}}

But, {{{m}}} < {{{CQP}}} + {{{m}}} < {{{CQN}}} + {{{m}}} < {{{BQN = 180}}}°

&#8658; {{{2m}}} < {{{CQN }}}= {{{180 - 62 = 118}}}°

&#8658; {{{m}}} < {{{CQN = 59}}}°


But, {{{m}}} <{{{BCQ}}} = {{{90}}} - {{{m}}} < {{{CQN }}}= {{{90 - 59 = 31}}}°

{{{m}}} < {{{PCQ}}} = {{{180}}}° -{{{ m}}} < {{{CPQ}}} - {{{m}}} < {{{CQP}}} = {{{180 - 78 - 62 = 40}}}°

Therefore, {{{m}}} < {{{ACB}}} = {{{m}}} < {{{ACP }}}+ {{{m}}} < {{{PCQ }}}+ {{{m}}} < {{{BCQ }}}= {{{39 + 40 + 31 = 110}}}°



so, your answer is:


{{{m}}} < {{{ACB}}} ={{{110}}}°