Chapter 12.6, Problem 25E

To determine

To conclude: That the $\Delta ABC\cong \Delta DBC$ from the given figure.

Answer to Problem 25E

It is proved that $\Delta ABC\cong \Delta DBC$ .

Explanation of Solution

Given:

The given figure is as follows.

  

Calculation:

Find the values of $x$ and $y$ .

By the Triangle SUM Property for $\Delta ABC$ .

  $\begin{array}{c}\angle ABC+\angle BCA+\angle CAB=180\\ \left(8x-32\right)+\left(5x+10\right)+\left(2x-8\right)=180\\ 15x-30=180\\ x=14°\end{array}$

So,

  $\begin{array}{c}\angle ABC=\left(8\cdot 14-32\right)\\ =80°\\ \angle BCA=\left(5\cdot 14+10\right)\\ =80°\end{array}$

And,

  $\begin{array}{c}\angle CAB=\left(2\cdot 14-8\right)\\ =20°\end{array}$

By the Triangle sum Property for $\Delta DBC$ .

  $\begin{array}{c}\angle DBC+\angle BCD+\angle CDB=180\\ \left(4y-24\right)+\left(3y+2\right)+\left(y-6\right)=180\\ 8y-28=180\\ y=26°\end{array}$

So,

  $\begin{array}{c}\angle DBC=\left(4\cdot 26-24\right)\\ =80°\\ \angle BCD=\left(3\cdot 26+2\right)\\ =80°\end{array}$

And,

  $\begin{array}{c}\angle CDB=\left(26-6\right)\\ =20°\end{array}$

From the above it is observed that $\angle ABC\cong \angle DBC$ , $\overline{BC}\cong \overline{BC}$ , and $\angle BCA\cong \angle BCD$ .

By ASA Congruence Theorem, $\Delta ABC\cong \Delta DBC$ .

From the above it is observed that $\angle BAC\cong \angle BDC$ , $\overline{BC}\cong \overline{BC}$ , and $\angle BCA\cong \angle BCD$ .

By AAS Congruence Theorem, $\Delta ABC\cong \Delta DBC$ .

From the above it is observed that $\angle BAC\cong \angle BDC$ , $\overline{BC}\cong \overline{BC}$ , and $\angle ABC\cong \angle DBC$ .

By AAS Congruence Theorem, $\Delta ABC\cong \Delta DBC$ .

Therefore, it is proved that $\Delta ABC\cong \Delta DBC$ .