Question

do it and attach then send plz

Answer 1

Answer:

$in \: tri \: (ehf) \: and \: tri(ghf) \\ ef = gf (given)\\ eh = gh(hf is bisector of eg) \\ fh = fh(common) \\ by \: sss \: congruency \\ angle \: efh \: = angle \: gfh (by \: cpct) \\ so \: hf \: is \: angle \: bisector \: of \: angle \: efg$

Step-by-step explanation:

mark as brainliest pl

have a good day