Question

In Fig. 37, AD bisects ∠A and AD ​⊥ BC.
(i) Is ∆ ADB ≅ ∆ ADC?
(ii) State the three pairs of matching parts you have used in (i).
(iii) Is it true to say that BD = DC?

Answer 1

$$\begin{gathered} \text{(i) Yes, △ADB}\cong \text{△ADC, by ASA criterion of congruency} \\ \text{(ii)}\text{We have used}\text{∠BAD}\mathit{}\mathit{=}\mathit{\text{∠}}\text{CAD} \\ \text{∠ADB=∠ADC = 90° since AD⊥BC} \\ \text{and AD = DA} \\ \text{(iii) Yes, BD\hspace{0.17em}= DC since, △ADB ≅△ADC} \end{gathered}$$