Lesson 3.
1 Secondary Parts of a Triangle
Concurrent Lines are lines that intersect in a single point Secondary Parts of a Triangle:
Median of a Triangle a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side
Angle Bisector of a Triangle a segment which bisects an angle of a triangle and whoso endpoints are a vertex and a point on the opposite side
Altitude of a Triangle a segment from the vertex of the triangle perpendicular to the line containing the opposite side
Perpendicular Bisector of a Side of a Triangle a line whose points are equidistant from the endpoints of the given side
Names of Point of Concurrency Incenter the point of concurrency of the three angle bisector of the triangle
Centroid the point of concurrency of the three medians of the triangle
Orthocenter the point of concurrency of the three altitudes of the triangle
Circumcenter the point of concurrency of the three perpendicular bisectors of the sides of the triangle.
Concurrency Properties The circumcenter is equidistant from the vertice of the triangle The incenter is equidistant from the three sides of the triangle The centroid is two-thirds of the distance from each vertex to the midpoint of the opposite side.
Lesson 3.2 Midsegment
Midsegment Theorem The segment joining the midpoints of the two sides of a triangle is parallel to the third side and half as long as the length of the third side. Midsegment Formula: CD= OA
Segment DE is the midsegment of triangle CBA Example:
In triangle VAL, m = 90 and points E and N are midpoints of segment VA and segment LA, respectively. Find a.mANE b. mAEN c.mVEN a. L and ANE are corresponding angles. Since segment EN||segment VL, m L = 90. b. The sum of the measures of the angles of a triangle is 180, therefore, mAEN = 180 (mANE + mA) = 180 (90+51) = 39 c. Since AEN and VEN form a linear pair, then mVEN = 180 - mAEN = 180 39 = 141
Lesson 3.3 Congruent Figures
Congruent Triangles (CPCTC) two triangles are congruent if and only if their corresponding parts are congruent. Example:
Sides: Segment AB Segment BC Segment AC Segment DE Segment EF Segment DF
Angles: Angle A Angle B Angle C Angle D Angle E Angle F
Therefore, Triangle ABC is congruent to Triangle DEF.
Lesson 3.4 Proving Triangles Congruent (The SSS Postulate)
The SSS Postulate If three sides of one triangle are congruent to the corresponding sides of the other triangle, then the two triangles are congruent. Example:
Given : P is the midpoint of segment BC and segment AD : segment AB Prove: Triangle APD Proof: P is the midpoint of segment AD ----- Given Segment AP Segment DP ----- Def of Midpoint segment CD Triangle BPC
P is the midpoint of segment BC ----- Given Segment DP Segment AB Triangle ABP Segment CP ----- Def of Midpoint Segment CD ----- Given Triangle DCP ----- Def of Midpoint
Lesson 3.5 The SAS and ASA Congruent Postulates
The SAS Postulate If two sides and the included angle of one triangle are congruent to the corresponding two sides and the included angle of another triangle, then the two triangles are congruent. Example:
If segment AB .:. Triangle ABC
segment PQ, B
Q, and segment BC
segment QR,
PQR by the SAS Postulate
The ASA Postulate If two angles and the included side of one triangle are congruent to the corresponding two angles and the included side of another triangle, then the two triangles are congruent. Example:
They are congruent because of the ASA postulate.
Perpendicular Bisector Theorem If a point lies on the perpendicular bisector of a segment, then the point is equidistant from the endpoints of the segment.
Given: l is the perpendicular bisector of segment AB and P is on l.
