Similarity, SMB-023

Show that \(\Delta ACD\) and \(\Delta DCB\) in the figure below are similar. (2 marks)

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\(\text{Proof (See Worked Solution)}\)

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\(\angle BDC = 180-(88+44) = 48^{\circ} \ \text{(angle sum of Δ)} \)

\(\angle CAD = \angle CDB = 48^{\circ} \)

\(\angle ACD=44^{\circ} \ \text{is common to}\ \Delta ACD\ \text{and}\ \Delta BCD \)

\(\therefore \Delta ACD\ \text{|||}\ \Delta DCB\ \ \text{(equiangular)}\)