Question

In each of the following pairs of right triangles, the measures of some parts are indicated along side. State by the application of RHS congruence condition which are congruent. State each result in symbolic form. (Fig. 46)

Answer 1

i) $$\begin{gathered} \text{∠ADC}=\text{∠BCA}=90° \\ \text{AD}=\text{BC} \\ \text{and hyp AB}=\text{hyp AB} \\ \text{Therefore, by RHS,}△\text{ADB}\cong △\text{ACB.} \end{gathered}$$

ii)

$$\begin{gathered} \text{AD=AD}\text{(Common)} \\ \text{hyp AC}=\text{hyp AB}\text{(Given)} \\ \text{∠ADB}+\text{∠ADC}=180°\text{(Linear pair)} \\ \angle \text{ADB}+90°=180° \\ \angle \text{ADB}=180°-90°=90° \\ \angle \text{ADB}=\angle \text{ADC}=90° \\ \text{Therefore,}\text{by RHS,}\text{△ADB}=\text{△ADC} \end{gathered}$$

iii)
$$\begin{gathered} \text{hyp AO}=\text{hyp DO} \\ \text{BO}=\text{CO} \\ \angle \text{B}=\angle \text{C}=90° \\ \text{Therefore,}\text{by RHS,}\text{△AOB}\cong \text{△DOC} \end{gathered}$$

iv)
$$\begin{gathered} \text{Hyp AC}=\text{Hyp CA} \\ \text{BC}\hspace{0.17em}=\text{DC} \\ \text{∠ABC}=\text{∠ADC}=90° \\ \text{Therefore,}\text{by RHS,}\text{△ABC}\cong \text{△ADC} \end{gathered}$$

v)
$$\begin{gathered} \text{BD}=\text{DB} \\ \text{Hyp AB}=\text{Hyp BC,}\text{as per the given figure.} \\ \text{∠BDA}+\text{∠BDC}=180° \\ \text{∠BDA}+90°=180° \\ \text{∠BDA}=180°-90°=90° \\ \text{∠BDA}=\text{∠BDC}=90° \\ \text{Therefore,}\text{by RHS,}\text{△ABD}\cong \text{△CBD} \end{gathered}$$