Question

In Fig. 47, BD and CE are altitudes of ∆ ABC and BD = CE.
(i) Is ∆ BCD ≅ ∆ CBE?
(ii) State the three pairs of matching parts you have used to answer (i).

Answer 1

(i) Yes, $△BCD\cong △CBE$ by RHS congruence condition.
(ii) We have used hyp BC = hyp CB
BD = CE (given in question)
and $\angle BDC=\angle CEB=90°$.