Chapter 4.3, Problem 15WE

To determine

To prove: The angles made by the wired from the tree to the ground are congruent.

Given Information

A tree, PT, on level ground which is supported at ‘P’ by three wires of equal length.

So, according to question,

$\overline{\text{PA}}\cong \overline{\text{PB}}\cong \overline{\text{PC}}$

and

$\overline{\text{AT}}\cong \overline{\text{TB}}\cong \overline{\text{TC}}$

Proof

Now, considering, $\Delta \text{ATP and }\Delta \text{PTB}$ ,

$\overline{\text{AP}}\cong \overline{\text{PB}}$   (given)

$\overline{\text{AT}}\cong \overline{\text{TB}}$   (given)

$\overline{\text{PT}}\cong \overline{\text{TP}}$   (common)

So, by SSS congruency,

$\Delta \text{ATP}\cong \Delta \text{PTB}$

Hence, $\angle \text{PAT}\cong \angle \text{TBP}$ ...(i)

by congruent part of congruent triangles

Again, considering, $\Delta \text{PTB and }\Delta \text{PTC}$

$\overline{\text{PB}}\cong \overline{\text{PC}}$   (given)

$\overline{\text{TB}}\cong \overline{\text{TC}}$   (given)

$\overline{\text{PT}}\cong \overline{\text{TP}}$   (common)

So, by SSS congruency,

$\Delta \text{PTB}\cong \Delta \text{PTC}$

Hence, $\angle \text{PBT}\cong \angle \text{PCT}$ ...(ii)

by congruent part of congruent triangles

Conclusion

So, from (i) and (ii), we can conclude that

$\angle \text{PAT}\cong \angle \text{PBT}\cong \angle \text{PCT}$

Explanation of Solution

Given Information

A tree, PT, on level ground which is supported at ‘P’ by three wires of equal length.

So, according to question,

$\overline{\text{PA}}\cong \overline{\text{PB}}\cong \overline{\text{PC}}$

and

$\overline{\text{AT}}\cong \overline{\text{TB}}\cong \overline{\text{TC}}$

Proof

Now, considering, $\Delta \text{ATP and }\Delta \text{PTB}$ ,

$\overline{\text{AP}}\cong \overline{\text{PB}}$   (given)

$\overline{\text{AT}}\cong \overline{\text{TB}}$   (given)

$\overline{\text{PT}}\cong \overline{\text{TP}}$   (common)

So, by SSS congruency,

$\Delta \text{ATP}\cong \Delta \text{PTB}$

Hence, $\angle \text{PAT}\cong \angle \text{TBP}$ ...(i)

by congruent part of congruent triangles

Again, considering, $\Delta \text{PTB and }\Delta \text{PTC}$

$\overline{\text{PB}}\cong \overline{\text{PC}}$   (given)

$\overline{\text{TB}}\cong \overline{\text{TC}}$   (given)

$\overline{\text{PT}}\cong \overline{\text{TP}}$   (common)

So, by SSS congruency,

$\Delta \text{PTB}\cong \Delta \text{PTC}$

Hence, $\angle \text{PBT}\cong \angle \text{PCT}$ ...(ii)

by congruent part of congruent triangles

Conclusion

So, from (i) and (ii), we can conclude that

$\angle \text{PAT}\cong \angle \text{PBT}\cong \angle \text{PCT}$