6. In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Solution: It is given in the question that AB = AD, AC = AE, and ∠BAD = ∠EAC To prove: The line segment BC and DE are similar i.e. BC = DE Proof: We know that ∠BAD = ∠EAC Now, by adding ∠DAC on both sides we get, ∠BAD + ∠DAC = ∠EAC +∠DAC This implies, ∠BAC = ∠EAD Now, ΔABC and ΔADE are similar by SAS congruency since: (i) AC = AE (As given in the question) (ii) ∠BAC = ∠EAD (iii) AB = AD (It is also given in the question) ∴ Triangles ABC and ADE are similar i.e. ΔABC ≅ ΔADE. So, by the rule of CPCT, it can be said that BC = DE.
