Chapter 6.1, Problem 6WE

To determine

To Proof : For the given congruent triangles, $AK>RS$.

Explanation of Solution

Given information:

  $\Delta ABC\cong \Delta RST$

  

  $\Delta ABC\cong \Delta RST$ is already given

Hence, the reason to 1st statement is given.

Corresponding parts of congruent triangles are congruent.

Therefore,

  $\overline{AB}\cong \overline{RS}$

Hence, the reason to 2nd statement is Corresponding parts of congruent triangles are congruent.

From Segment Addition Postulate.

B is a point between A and K

Therefore,

  $AK=AB+BK$

Hence, the reason to 3rd statement is Segment Addition Postulate.

From Property of inequality.

If $a=b+c$ and $c>0$ then $a>b$

Therefore,

  $AK>AB$

Hence, the answer to 4 is Property of Inequality if a= b+ c and c > 0 then a > b.

Consider that

From 3rd statement ;

  $AK=AB+BK$

From 2nd statement;

  $\overline{AB}\cong \overline{RS}$

Therefore,

  $AK=RS+BK$

From property of inequality;

If $a=b+c$ and $c>0$ then $a>b$

Therefore,

  $AK>RS$

Hence, the answer to 5 is Property of Inequality If $a=b+c$ and $c>0$ then $a>b$ ..