Right Triangle In Triangle

Refer to the above diagram. Now we extend $\overline {BD}$ to $\overline {AC}$ and label the intersection point $B'$ . As $\angle BDA={ 90 }^{ o }$ and $\overline {AD}$ is the angle bisector of $\angle BAC$ , we can conclude that $\Delta BAB'$ is isosceles. This means that $|\overline {AB}|=|\overline {AB'}|=17$ .

Also, another property of an isosceles triangle is that the perpendicular bisector of the base is the median of the base from the opposite vertex, thus implying that $|\overline {BD}|=|\overline {DB'}|$ and hence $D$ is the midpoint of $\overline {BB'}$ . Furthermore, we are also given that $|\overline {BM}|=|\overline {MC}|$ , which means $M$ is the midpoint of $\overline {BC}$ . Hence we can apply midsegment theorem, and thus, $|\overline {DM}|= \frac{|\overline {B'C}|}{2} = \frac {61-17}{2} = 22$

Since,I am very poor with latex,I have only provided an outline of my solution.In the solution $a,b,c,A,B,C$ of triangle ABC have their usual meaning.Let us take the origin at point M.Suppose point C is denoted by complex number x , then point B will be denoted by complex number -x.Let y denote complex number corresponding to point A and d denote complex number corresponding to point D.I shall use the notation cis(k)=cos (k)+isin (k) for any angle k.Now, in triangle ABD, anticlockwise rotation about point D gives, (d+x)=(d-y)tan (A/2)cis (pi/2) (this is equation 1).Anticlockwise rotation in triangle ABD about point A gives, (y-d)=(y+x)cos (A/2)cis (A/2) ( this is equation 2).Now,we take the value of tan (A/2) from equation 1, and the value of sec (A/2) =1/(cos (A/2)) from equation 2, and using (sec (A/2))^2-(tan (A/2))^2=1,we can eliminate A and we get the equation 2d=(y-x)-(y+x)cis (A).Now in triangle ABC,anticlockwise rotation about point A gives, (y-x)=(y+x)(b/c)cis (A).Substituting the value of (y+x)cis (A) in the previous equation, we get 2d=(y-x)((b-c)/b). (This equation tells us that MD is parallel to CA).Taking modulus in both sides of equation we get, 2|d|=|(y-x)|((b-c)/b)=((b-c)/b)(b)=(b-c).Hence,we get |d|=(b-c)/2.But |d|=MD.Therefore, MD = (b-c)/2.

Okay, here is a "solution". Since $BC$ is not given to us, so we will take it to be $44$ , so the triangle is degenerate. Then, observe the following:

1) $A,B,C$ lie on a straight line in that order.

2) $D$ becomes $B$ .

3) $M$ is the midpoint of $BC$ , so $DM = BM = \frac {BC}{2} =22$ . HOORAY!

Note: I also solved the problem properly.