1. ## Triangle PQR...

In triangle PQR, <p = 30 degrees
<Q = 80 degrees
<R = 70 degreees
QH is an altitude and RM is a median and H is on the line PR,
the measure of <MHP is:

a) 15 degrees or 20 degrees or 30 degrees or 40 degrees or 45 degreess ?

diagram is attatched

2. Multiple hellos. It is 30 degrees.

Solution: (All angular measurements in degrees)
$\displaystyle \angle P = 30$ degrees
$\displaystyle \angle Q = 80$degrees
$\displaystyle \angle R = 70$ degrees
$\displaystyle QH$is an altitude and $\displaystyle RM$is a median and $\displaystyle H$ is on the line $\displaystyle PR$

If $\displaystyle QH$is an altitude, then $\displaystyle \angle QHR$ and $\displaystyle \angle QHP$ are $\displaystyle 90$.
Filling in relevant angles angles: $\displaystyle \angle PHQ=60$,

Area of $\displaystyle \triangle PQR=\frac{PRQH}{2}$

A median splits a triangle into two triangles equal in area.
Construct altitude $\displaystyle MO$. (dotted line in diagram, forgot to label O)

Two similar triangles are discovered, $\displaystyle \triangle MOP$ and $\displaystyle \triangle PHQ$ (as $\displaystyle MO$and $\displaystyle QH$are parallel)
Calculate area of $\displaystyle \triangle PMR=\frac{MOPR}{2}=\frac{PRQH}{4}$
(Because a median splits a triangle into two triangles equal in area.)

Hence,$\displaystyle MO=\frac{QH}{2}$
Which means that $\displaystyle PO=\frac{PH}{2}$ and $\displaystyle PM=\frac{PQ}{2}$

Thus $\displaystyle OH=PH-\frac{PH}{2}=\frac{PH}{2}$
Hence $\displaystyle \triangle MPO$ and$\displaystyle \triangle MOH$ are identical
Hence $\displaystyle \angle MHP=30$