Chapter 3.4, Problem 10PSB

To determine

To prove: AWand BW line segments in the given triangle are congruent.

Explanation of Solution

Given information:

In a given triangle WXY,

  $\begin{array}{l}\overline{\text{AZ}}\cong \overline{\text{ZB}}\\ \text{Z is the midpoint of }\overline{\text{XY}}\\ \angle \text{AZX}\cong \angle \text{BZY}\\ \overline{\text{XW}}\cong \overline{\text{YW}}\end{array}$

Formula used:

Congruent Parts of Congruent Triangles are Congruent (CPCTC)

By definition of congruency

SAS criteria of congruency.

Proof:

  $\begin{array}{l}\text{In }\Delta \text{s AXZ and BYZ, we have}\\ \overline{\text{AZ}}\cong \overline{\text{BZ}}\mathrm{..........}Given\\ \therefore \overline{\text{AZ}}=\overline{\text{BZ}}\mathrm{.......}\text{by definition of congruency}\mathrm{....}\left(i\right)\\ Z\text{ is the mid point of }\overline{\text{XY}}\mathrm{......}Given\\ \therefore \overline{\text{XZ}}=\overline{\text{YZ}}\mathrm{.............}\left(ii\right)\\ \angle \text{AZX}\cong \angle \text{BZY}\mathrm{.......}\text{Given}\\ \therefore \angle \text{AZX=}\angle \text{BZY}\mathrm{........}\left(iii\right)\\ \text{From }\left(i\right),\left(ii\right)\text{ and }\left(iii\right),\text{we have}\\ \Delta \text{ AXZ }\cong \Delta \text{BYZ}\\ \angle \text{ AXZ }\cong \angle \text{BYZ}\mathrm{..........}\text{CPCTC}\\ \overline{\text{XA}}=\overline{\text{YB}}\mathrm{.........}\text{CPCTC}\\ \text{Now, join W to Z}\end{array}$

  $\begin{array}{l}\text{In }\Delta \text{WXZ and }\Delta \text{WYZ}\\ \overline{\text{XZ}}=\overline{\text{YZ}}\mathrm{................}\left(a\right)\\ \angle \text{ AXZ }\cong \angle \text{BYZ}\mathrm{..........}\text{CPCTC}\\ \angle \text{ AXZ = }\angle \text{BYZ}\mathrm{..........}\text{by definition}\\ \therefore \angle \text{WXZ=}\angle \text{WYZ}\mathrm{.....}\left(b\right)\\ \angle \text{ WZX = }\angle \text{WZY=90}°\mathrm{.....}\left(c\right)\\ \text{From }\left(a\right),\left(b\right)\text{ and }\left(c\right),\text{we have}\\ \Delta \text{WXZ }\cong \Delta \text{WYZ}\\ \overline{\text{WX}}\cong \overline{\text{WY}}\mathrm{........}\text{Given}\\ \therefore \overline{\text{WX}}-\overline{\text{AX}}\cong \overline{\text{WY}}-\overline{\text{BY}}\\ \text{⇒}\overline{\text{AW}}\cong \overline{\text{BW}}\\ \text{Hence proved}\end{array}$