Question

problem 24: given: $overline{cd}perpoverline{ad}$; $overline{bc}perpoverline{ab}$; $overline{ac}$ bisects $angle dcb$; $angle1congangle3$; $angle2congangle4$. prove: $angle5congangle6$

Explanation:

Step1: Use angle - bisector property

Since $\overline{AC}$ bisects $\angle DCB$, we have $\angle 1=\angle 2$.

Step2: Consider right - angled triangles

In right - triangle $ADC$, $\angle 1+\angle 5 = 90^{\circ}$ (because $\overline{CD}\perp\overline{AD}$). In right - triangle $ABC$, $\angle 2+\angle 6=90^{\circ}$ (because $\overline{BC}\perp\overline{AB}$).

Step3: Substitute equal angles

Since $\angle 1 = \angle 2$, we can substitute $\angle 1$ for $\angle 2$ in the equations. So, $\angle 1+\angle 5=\angle 2+\angle 6$. Subtracting $\angle 1$ (which is equal to $\angle 2$) from both sides of the equation, we get $\angle 5=\angle 6$, or $\angle 5\cong\angle 6$.

Answer:

$\angle5\cong\angle6$ is proved.