Answer:
[tex]m \angle A = m \angle C[/tex] by reason [tex]\overline{AB} \cong \overline{BC}[/tex] and [tex]m \angle B = m \angle M = m \angle P[/tex].
[tex]\triangle AMO \cong \triangle CPO[/tex] SAS Theorem
Step-by-step explanation:
We proceed to demonstrate the statement by Geometric means:
1) [tex]\overline{AB} \cong \overline{BC}[/tex], [tex]\overline {AM} \cong \overline {PC}[/tex], [tex]m\angle AMO = m\angle CPO[/tex] Given.
2) [tex]\frac{AM}{AB} = \frac{PC}{BC}[/tex] Proportionality.
3) [tex]\frac{AM}{AM + MB} = \frac{PC}{BP + PC}[/tex] Definition of line segments.
4) [tex]\frac{1}{1+\frac{MB}{AM} } = \frac{1}{\frac{BP}{PC}+1}[/tex] Algebra.
5) [tex]\frac{BP}{PC} + 1 = 1 +\frac{MB}{AM}[/tex] Algebra.
6) [tex]\frac{BP}{PC} = \frac{MB}{AM}[/tex] Algebra.
7) [tex]BP = BM[/tex] By 1)
8) [tex]m \angle B = m \angle M = m \angle P[/tex] By 1), 7)
9) [tex]\triangle AMO \sim \triangle ABC[/tex], [tex]\triangle CPO \sim \triangle ABC[/tex] By 1), 7), 8). Defintion of simmilarity.
10) [tex]\frac{AM}{MO} = \frac{AB}{BC}[/tex], [tex]\frac{PO}{PC} = \frac{AB}{BC}[/tex] Definition of proportionality.
11) [tex]\frac{AM}{MO} = \frac{PO}{PC}[/tex] Algebra.
12) [tex]AM^{2} = PO\cdot MO[/tex] Algebra.
13) [tex]PO = MO[/tex] By 12) and Algebra.
14) [tex]\overline{PO} \cong \overline{MO}[/tex] By 13).
15) [tex]\triangle AMO \cong \triangle CPO[/tex] SAS Theorem/Result.
