Chapter 10.1, Problem 10WE

To determine

To prove: that $\Delta OBT\cong \Delta EFG$

Explanation of Solution

Calculation:

Construct the triangle $\Delta OBT$

Construction:

A triangle congruent to $\Delta OBT$ using ASA postulate,

  

Draw a line $L$ .

Mark a point $F$ on the line $L$ . Taking $F$ as center draw an arc on the line $L$ with radius of $BT$ .

With $B$ as center draw an arc which touches the sides $BO$ and $BT$ .

With same radius draw an arc with $F$ as center.

Then mark the distance between the points $P$ and $Q$ . Draw an arc of same measure with $S$ as center.

Draw a ray at the point of intersection, marked it as $J$ .

With $T$ as center, draw an arc which touches the sides $TB$ and $TO$ . With same radius draw an arc with $G$ as center.

Then mark the distance between the points $U$ and $V$ . Draw an arc of same measure with $X$ as center. Draw a ray at the point of intersection, marked it as $K$ .

Then mark the point of intersection as $E$ .

  

Angle-Side-Angle theorem states that if two angles and any side of one triangle are congruent to two angles and any side of another triangle, then the triangles are congruent.

Thus, $\Delta OBT$ is congruent to $\Delta EFG$ .

Justification: Since $\angle B\cong \angle F,\angle T\cong \angle G$ and $BT\cong FG$ then $\Delta OBT\cong \Delta EFG$ .