Question

perpendicu
Hence every point on the perpendicular
the end points of the segment.
Part II : Any point equidistant from the end points of a segment lies on the
bisector of the segment.
Given : Point P is any point equidistant from
the end points of seg AB. That is, PA = PB.
To prove: Point P is on the perpendicular bisector of seg AB.
Construction : Take mid-point M of seg AB and draw line PM.
Proof : In A PAM and A PBM
Р P.
H
+
#
A
M M
seg PA S seg PB
seg AM S seg BM .......
B
seg PMS
common side
..A PAMEA PBM ......
test.
.. ZPMA ZPMB.......c.a.c.t.
Fig. 3.41
But ZPMA +
= 180°
ZPMA + ZPMA = 180° ........ ( ZPMB = ZPMA)
2 ZPMA =
.. ZPMA = 90°
.. seg PMI seg AB
...... (1)
But Point M is the midpoint of seg AB. ......construction .... (2)
.. line PM is the perpendicular bisector of seg AB. So point P is ont
perpendicular bisector of seg AB​

Answer 1

Answer:

I cant understand the question

Answer 2

Answer:

given

given

mp

sss test

pmb

180°