Chapter 4.8, Problem 5CYU

To determine

To Prove:

  $△FGH\cong △FDC$

Explanation of Solution

Give information:

  

Proof:

If three sides of one triangle to three sides of a second triangle, then the triangles are congruent.

Find the length of $\overline{GH}$ with $G\left(a,b\right)$ and $H\left(a,0\right)$ ,

  $\begin{array}{l}GH=\sqrt{{\left(x_2-x_1\right)}^2-{\left(y_2-y_1\right)}^2}\\ GH=\sqrt{0+b^2}\\ GH=b\\ GH=\sqrt{{\left(a-a\right)}^2+{\left(b-0\right)}^2}\end{array}$

Find the length of $\overline{GF}$ with $G\left(a,b\right)$ and $F\left(0,\frac{b}{2}\right)$

  $\begin{array}{l}GF=\sqrt{{\left(x_2-x_1\right)}^2-{\left(y_2-y_1\right)}^2}\\ GF=\sqrt{{\left(a-0\right)}^2+{\left(b-\frac{b}{2}\right)}^2}\\ GF=\sqrt{a^2+\frac{b^2}{4}}\end{array}$

Find the length of $\overline{HF}$ with $H\left(a,0\right)$ and $F\left(0,\frac{b}{2}\right)$

  $\begin{array}{l}HF=\sqrt{{\left(x_2-x_1\right)}^2-{\left(y_2-y_1\right)}^2}\\ HF=\sqrt{{\left(a-0\right)}^2+{\left(0-\frac{b}{2}\right)}^2}\\ HF=\sqrt{a^2+\frac{b^2}{4}}\end{array}$

After that find out length of each side $△FDC$ ,

Length of $\overline{DC}$ with $D\left(-a,b\right)$ and $C\left(-a,0\right)$

  $\begin{array}{l}DC=\sqrt{{\left(x_2-x_1\right)}^2-{\left(y_2-y_1\right)}^2}\\ DC=\sqrt{{\left(-a-\left(-a\right)\right)}^2+{\left(b-0\right)}^2}\\ DC=\sqrt{0+b^2}\\ DC=b\end{array}$

To find the Length of $\overline{DF}$ with $D\left(-a,b\right)$ and $F\left(0,\frac{b}{2}\right)$

  $\begin{array}{l}DF=\sqrt{{\left(x_2-x_1\right)}^2-{\left(y_2-y_1\right)}^2}\\ DF=\sqrt{{\left(0-\left(-a\right)\right)}^2+{\left(\frac{b}{2}-b\right)}^2}\\ DF=\sqrt{a^2+\frac{b^2}{4}}\end{array}$

To find the Length of $\overline{CF}$ with $C\left(-a,0\right)$ and $F\left(0,\frac{b}{2}\right)$

  $\begin{array}{l}CF=\sqrt{{\left(x_2-x_1\right)}^2-{\left(y_2-y_1\right)}^2}\\ CF=\sqrt{{\left(0-\left(-a\right)\right)}^2+{\left(\frac{b}{2}-0\right)}^2}\\ CF=\sqrt{a^2+\frac{b^2}{4}}\end{array}$

Thus, $\overline{GH}\cong \overline{DC},\overline{GF}\cong \overline{DF},\overline{HF}\cong \overline{CF}$ .