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Abcdefgh is a regular octagon. M is a point on the line dh. N is a point on the line fm. The line dn and fm are perpendicular. Prove that triangles fhm and dfp are congruent

Abcdefgh is a regular octagon M is a point on the line dh N is a point on the line fm The line dn and fm are perpendicular Prove that triangles fhm and dfp are class=

Respuesta :

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

Methods used to prove that the two triangles are congruent

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;


  • ΔFHM ≅ ΔDFP by definition of congruency

Learn more about the properties of a regular octagon here:

https://brainly.com/question/4503941

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