<PRE><B>Question 1177800</B>
.


&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;The solution by &amp;nbsp;@MathLover1 is &amp;nbsp;INCORRECT.


&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;I came to bring the correct solution.



&lt;pre&gt;
Let the given triangle be ABC, with the right angle B;

     the leg AB = 12;

         leg BC =  9,

     and the hypotenuse AC = 15.


The bisector AD of the acute angle A divides the leg BC in two parts BD and DC.


Then the ratio   |BD|  to  |CD|  is equal to the ratio  |AB|  to  |AC|,  i.e.  {{{12/15}}} = {{{4/5}}}.


It implies that  |BD| = 4,  |BC| = 5.



Hence,  the length of the bisector AD is equal to  {{{sqrt(12^2+4^2)}}} = {{{4*sqrt(3^2+1^1)}}} = {{{4*sqrt(10)}}}.    &lt;U&gt;ANSWER&lt;/U&gt;
&lt;/pre&gt;

Solved &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(C O R R E C T L Y).



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Looking into her activity in last several days, I start thinking that


@MathLover1 presents a real danger for any visitor to this forum, because she provides wrong solutions even to simplest Math problems.



See my notes to her posts in past two days


https://www.algebra.com/algebra/homework/Finance/Finance.faq.question.1177850.html


https://www.algebra.com/algebra/homework/Volume/Volume.faq.question.1177847.html


https://www.algebra.com/algebra/homework/Surface-area/Surface-area.faq.question.1177846.html


https://www.algebra.com/algebra/homework/Triangles/Triangles.faq.question.1177800.html


https://www.algebra.com/algebra/homework/Triangles/Triangles.faq.question.1177804.html



</PRE>