Chapter 3.4, Problem 7PSA

To determine

To prove:Two exterior angles in given triangle are congruent

Explanation of Solution

Given information:

  $\overline{\text{KP}}\text{ is a median of the given triangle and}$

  $\overline{\text{MK}}\cong \overline{\text{RK}}$

Formula used:

Median in a triangle divides the side in two equal parts

Congruent parts of congruent triangles are congruent (CPCTC)

Sum of angles formed on same side of straight line is $\text{180}°$

Proof :

  $\begin{array}{l}\overline{\text{KP}}\text{ is a median of given triangle}\\ \text{⇒}\overline{\text{RP}}=\overline{\text{PM}}\\ \overline{\text{MK}}\cong \overline{\text{RK}}\mathrm{.......}\text{Given}\\ \overline{\text{MK}}=\overline{\text{RK}}\mathrm{.......}\text{by definition of congruency}\\ \text{PK = PK, common side of both the triangle}\\ \therefore \Delta \text{ KPR }\cong \Delta \text{ KPM}\\ \text{and by CPCTC,}\\ \angle \text{KRP = }\angle \text{KMP}\mathrm{........}\left(i\right)\\ \therefore \angle 3+\angle \text{KRP = }\angle \text{4 + }\angle \text{KMP = 180}°\\ \text{⇒}\angle 3\text{ = }\angle \text{4}\mathrm{.........}\text{since, from }\left(i\right),\angle \text{KRP = }\angle \text{KMP}\\ \text{⇒}\angle 3\cong \angle \text{4}\mathrm{.......}\text{ by definition of congruency}\\ \text{Hence proved}\end{array}$