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The bisector of the base angles Q and R of an equilateral triangle PQR meet at S. ST and SM are drawn parallel to the sides PQ and PR. Show that QT=TM=MR

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Answer:

From TM = MS, MS = MR, and TM = MR

QT = TM = MR Transitive property

Explanation:

ΔPQR is an equilateral triangle Given

∠PQR = ∠RPQ = ∠PRQ = 60° Definition of equilateral triangle

∠PQR = ∠SQT + ∠SQP = 60° Addition property of angles

∠PRQ = ∠SRM + ∠SRP = 60° Addition property of angles

In triangle PQR, ∠SQT = ∠SQP = 30° = ∠SRM = ∠SRP Angles bisected by angle bisector

∠STM = ∠PQR - 60° Angle of same side of two parallel lines crossing the same transversal

∠PRQ = ∠SMT = 60° Angle of same side of two parallel lines crossing the same transversal

∠MST = 180 - ∠SMT - ∠STM = 180 - 60 - 60 = 60°

∠MST = ∠SMT = ∠STM = 60°, ΔMST is an equilateral triangle, all angles equal

ST = MS = TM definition of equilateral triangle

∠STQ = 180 - ∠STM = 180° - 60° = 120° (∠STQ and ∠STM are supplementary angles)

∠QST = 180 - ∠SQT - ∠STQ = 180 - 30 - 120 = 30 (Interior angles in the same triangle)

ΔQTS is an isosceles triangle, base angles are equal

∴ ST = QT Definition of equilateral triangle

Similarly

∠SMR = 180 - ∠SMT = 180° - 60° = 120° ∠SMR and ∠SMT are supplementary angles

∠MSR = 180 - ∠SRM - ∠SMR = 180 - 30 - 120 = 30 (Angles in the same triangle)

ΔRMS is an isosceles triangle, base angles are equal

∴ MS = MR Definition of equilateral triangle

Since ST = QT, and, ST = TM, QT = TM Transitive property

Also since TM = MS, and MS = MR, TM = MR Transitive property

Therefore;

QT = TM = MR Transitive property

The bisector of the base angles Q and R of an equilateral triangle PQR meet at S. ST-example-1
User Vasco Ferreira
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