91Ó°ÊÓ

The sum of the angles of a triangle is 180. The measure of an exterior angle of a triangle is greater than the measure of either remote interior angle.

Solve the sum of angles as:

$∠C+∠CAB+∠CBA=180°$ …… (1)

$∠K+∠KAB+∠KBA=180°$ …… (2)

By transitive property:

$∠C+∠CAB+∠CBA=∠K+∠KAB+∠KBA$ …… (a)

Since a whole part is equal to the sum of its parts as:

$∠CAB=∠CAK+∠KAB$ …… (3)

$∠CBA=∠CBK+∠KBA$ …… (4)

Substitute equation (3) and (4) into equation (a) as follows:

$\begin{array}{c}∠C+∠CAK+∠KAB+∠CBK+∠KBA=∠K+∠KAB+∠KBA\\ ∠C+∠CAK+∠CBK=∠K\end{array}$

Since a whole is greater than any of its part, therefore $m∠K>m∠C$.