Answer:
Option C. is correct
Step-by-step explanation:
In the triangle ABC, vertices are A(12, 8), B(4, 8) and C(4, 14)
AB = [tex]\sqrt{(12-4)^{2}+(8-8)x^{2}}=8[/tex]
BC = [tex]\sqrt{(4-4)^{2}+(8-14)^{2}}=6[/tex]
AC = [tex]\sqrt{(12-4)^{2}+(8-14)^{2})}=10[/tex]
In triangle XYZ vertices are X(6,6), Y(4,12) and Z(10, 14)
XY = [tex]\sqrt{(6-4)^{2}+(6-12)^{2}}=2\sqrt{10}[/tex]
YZ = [tex]\sqrt{(4-10)^{2}+(12-14)^{2}}=2\sqrt{10}[/tex]
XZ = [tex]\sqrt{(10-6)^{2}+(14-6)^{2}}=4\sqrt{5}[/tex]
In triangle MNO, vertices are M(4, 16), N(4, 8) and O(-2, 8)
MN = [tex]\sqrt{(4-4)^{2}+(16-8)^{2}}=8[/tex]
MO = [tex]\sqrt{(4+2)^{2}+(16-8)^{2}}=10[/tex]
NO = [tex]\sqrt{(4+2)^{2}+(8-8)^{2}}=6[/tex]
In triangle JKL, vertices are J(14, -2), K(12,2) and L(20, 4)
JK = [tex]\sqrt{(14-12)^{2}+(-2-2)^{2}}=2\sqrt{5}[/tex]
KL = [tex]\sqrt{(20-12)^{2}+(4-2)^{2}}=2\sqrt{17}[/tex]
JL = [tex]\sqrt{(20-14)^{2}+(4+2)^{2}}=6\sqrt{2}[/tex]
Now we compare the sides for the congruence
Since AB = MN = 8
BC = NO = 6
AC = OM = 10
So ΔABC ≅ Δ MNO
Option C. is correct
