Question 1155795


In triangle {{{ABC}}}, point {{{P}}} is on side {{{BC}}} such that

{{{ PA = 13}}}, 

{{{PB = 14}}}, 

{{{PC = 4}}}, 

and the circumcircle of triangles {{{APB}}} and {{{APC}}} have the same radius. Find the area of triangle {{{ABC}}}.


<a href="https://www.imageupload.net/image/a4e0s"><img src="https://imagehost.imageupload.net/2020/04/11/triangle-ABC.th.png" alt="triangle-ABC.png" border="0" /></a>


Note that  {{{AP}}} is a {{{chord}}} of both circles. Since both circles have the {{{same}}} {{{radius}}}, chord {{{AP}}}  must subtend the {{{same}}} angle, i.e. < {{{ABP}}}= < {{{ACP}}}. 

Thus, triangle {{{ABC}}} is {{{isosceles}}} with {{{AB=AC}}}.

Let{{{ M}}} be the {{{midpoint}}} of {{{BC}}}. We see that 

{{{BC=BP+ PC=14+4=18}}}

so{{{ BM=CM=BC/2=9}}} 

 
Then {{{PM=MC-PC=9-4=5}}}. 

Since {{{AM}}} is also an {{{altitude}}}, we can apply Pythagoras theorem to right triangle  {{{AMP}}} to get {{{AM}}}:

{{{AM=sqrt(13^2-5^2)=sqrt(169-25)=sqrt(144)=12}}}

Hence, the area of triangle {{{ABC}}} is:

{{{area=(1/2)AM*BC}}}

{{{area=(1/2)12*18}}}

{{{area=6*18}}}

{{{area=108}}}