Question

Which of the following pairs of triangles are congruent by ASA condition?

Answer 1

1) We have

$$\begin{gathered} \text{Since ∠ABO = ∠CDO}=45°\text{and both are alternate angles}, \\ \text{AB}\parallel \text{DC} \\ \text{∠BAO}=\text{∠DCO}(\text{alternate angle},\text{AB}\parallel \text{CD}\text{and AC is a transversal line)} \\ \text{∠ABO}=\text{∠CDO}=45°\text{(given in the figure)} \\ \text{Also, AB}=\text{DC}(\text{Given in the figure}) \\ \text{Therefore},\text{by ASA}△\text{AOB}\cong △\text{DOC} \end{gathered}$$

2)
$$\begin{gathered} \text{In △ABC ,} \\ \text{Now AB = AC (Given)} \\ \text{∠ABD = ∠ACD = 40° (Angles opposite to equal sides)} \\ \text{∠ABD +∠ACD+∠BAC=180° (Angle sum property)} \\ \text{40°+40°+∠BAC=180°} \\ \text{∠BAC=180°-80°=100°} \\ \text{∠BAD +∠DAC=∠BAC} \\ \text{∠BAD=∠BAC-∠DAC=100°-50°=50°} \\ \text{∠BAD =∠CAD = 50°} \\ \text{Therefore, by ASA, △ABD ≅△ADC} \end{gathered}$$
3)
$$\begin{gathered} \text{In∆ABC,} \\ \text{∠A+∠B+∠C=180°(Angle sum property)} \\ \text{∠C=180°-∠A-∠B} \\ \text{∠C=180°-30°-90°=60°} \\ \text{In∆PQR,} \\ \text{∠P+∠Q+∠R=180°(Angle sum property)} \\ \text{∠P=180°-∠Q-∠R} \\ \text{∠P=180°-60°-90°=30°} \\ \text{∠BAC = ∠QPR = 30°} \\ \text{∠BCA=∠PRQ = 60°} \\ \text{and AC = PR (Given)} \\ \text{Therefore, by ASA, △ABC ≅△PQR} \end{gathered}$$

4)
$$\begin{gathered} \text{We have only BC =QR but none of the angles of △ABC AND △PQR are equal.} \\ \text{Therefore, △ABC≇ △PRQ} \end{gathered}$$