Hello, mido22!
ABC is a triangle where AB = 5cm, AC = 3cm, D is the midpoint of BC.
Draw AH = 6√2 cm, perpendicular to both AC and AB.
Find the length of DH.
The diagram looks like this.
I don't see a unique length for $\displaystyle D\!H.$Code:H o | * *| _* |6√2 * * | * | * * o - - - - - o B *A 5 * * * * *3 o ** * D * * o C
It depends on $\displaystyle \theta = \angle B\!AC.$
If $\displaystyle \theta = 0^o,\:D\!H = 7\sqrt{2}$
If $\displaystyle \theta = 180^o,\:D\!H = 6\sqrt{2}$