Prove that in a right triangle $ABC$, the bisector of the right angle bisects the angle formed between the median and the altitude drawn from the same vertex.
In other words, I'm asked to prove that $\angle MBD = \angle DBH$; according to the picture below.
This is the same as proving that $\displaystyle \frac{BM}{BH} = \frac{MD}{DH}$
Others facts are also known:
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$AM = MB = MC$
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$\angle ABD = \angle CBD = 45º$
I'm not sure how to use that information. So what I tried to do is to construct a triangle so that I would be able to prove the equality by similarity, but I'm stuck:
$1)$ $DF \perp AB$, by construction.
$2)$ Quadrilateral $BFDH$ is circumscribed by a circle, since $m\angle BFD + m\angle BHD = 180º$.
$3)$ $\angle HFD = \angle DBH$, inscribed angles subtended by the same arc $\stackrel{\frown}{DH}$ (It follows from $2)$).
$4)$ $\angle FDB = \angle FHB$, inscribed angles subtended by the same arc $\stackrel{\frown}{FB}$ (It follows from $2)$).
$5)$ $\triangle FDO \sim \triangle BHO$, by AA similarity.
I don't know how to go further. It would be nice if there is some way to prove that $\angle BPD = \angle BOH$.
Best Answer
Here is the way:
$\angle ABM=\angle CAB$ (isocele triangle), and $\angle HBC=\angle CAB$ as well ($HBC$ $HAB$ similar triangles).
$\angle ABD=\angle DBC$ by definition, thus you have
$\angle DBM=\angle ABD-\angle ABM=\angle DBC-\angle HBC=\angle HBD$, which is what you wanted to prove.