Q28

 



Below is the correct solution but let me point out few things before that:

1. When you see 2 angles being formed by the same base somewhere in the problem, it's tempting to look for a cyclic quadrilateral.

2. Can we way that ABCD is a cyclic quadrilateral since the chord BD subtends the same angle at A and C? No. Because these angles are on the opposite sides of BD. What do I mean by opposite sides? If I extend BD infinitely in both directions, these angles are on the opposite sides of that.

3. So we have 2 equal angles in BAD and BCD but on opposite sides. How do we bring them to the same side? That's when we decided to extend CD to the other side and made a congruent triangle pair so that we can transfer this common angle on the same side.




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