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Question

# Question 7 O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that ΔOCD is an isosceles triangle.

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Solution

## O is a point in the interior of a square ABCD such that ΔOAB is an equilateral triangle. Construction Join OC and OD. AOB is an equilateral triangle. ∴∠OAB=∠OBA=60∘ ...(i) Also, ∠DAB=∠CBA=90∘...(ii) [each angle of a square is 90∘] [∵ ABCD is a square] On subtracting Eq. (i) from Eq. (ii) we get ∠DAB−∠OAB=∠CBA−∠OBA=90∘−60∘i.e., ∠DAO=∠CBO=30∘ In ΔAOD and ΔBOC, AO = BO [given] [all the side of an equilateral triangle are equal] ∠DAO=∠CBO [proved above] and AD = BC [sides of a square are equal] ∴ΔAOD≅ΔBOC [by SAS congruence rule] Hence, OD = OC [by CPCT] In ΔCOD, OC = OD Hence, ΔCOD is an isosceles triangle.

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