ΔFHM and ΔDFP by addition to ΔMOF and ΔPDO respectively gives
ΔHOP and ΔDOF, which are congruent.
Response:
- ΔFHM ≅ ΔDFP by transitive property of equality
<h3>Methods used to prove that the two triangles are congruent</h3>
Based on the properties of a regular octagon, the diagonals DH and BF
of a regular octagon are perpendicular, and they bisect each other,
given that the length of half the diagonal represent the radius of the
circumscribing circle.
Therefore;
∠NFP ≅ ∠NFP by reflexive property
∠PNF ≅ ∠MOF all right angles are congruent
ΔMOF is similar to ΔPNF by Angle-Angle AA similarity postulate.
∠OMF ≅ ∠DMN by reflexive property
Therefore;
ΔDMN is similar to ΔMOF by AA similarity postulate.
Similarly, we have;
ΔDMN ~ ΔPDO by AA similarity postulate;
Therefore;
ΔPDO is similar to ΔOMF by transitive property
∠ODP ≅ ∠OFM by CASTC, Corresponding Angles of Similar Triangle are Congruent.
OD = OF distance from center to the vertex of a regular octagon are
equal to the radial length of the circumscribing circle and are therefore
equal.
Therefore;
ΔPDO ≅ ΔOMF by Angle-Side-Angle ASA congruency rule
ΔPDO = ΔOMF definition of congruency
- ΔHOF ≅ ΔDOF by Side-Angle-Side, SAS, congruency rule
ΔHOF = ΔDOF
ΔFHM = ΔHOF - ΔOMF subtraction property
Similarly
ΔDFP = ΔDOF - ΔPDO = ΔHOF - ΔOMF = ΔFHM by substitution property
ΔDFP = ΔFHM
ΔFHM = ΔDFP by symmetric property.
Therefore;
<u>ΔFHM ≅ ΔDFP by definition of congruency</u>
Learn more about the properties of a regular octagon here:
brainly.com/question/4503941
