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)
$\angle P = 30$ degrees
$\angle Q = 80$degrees
$\angle R = 70$ degrees
$QH$is an altitude and $RM$is a median and $H$ is on the line $PR$

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

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

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

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

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

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