Question : ABC is an isosceles triangle where AB = AC, which is circumscribed about a circle. If P is the point where the circle touches the side BC, then which of the following is true?
Option 1: BP = PC
Option 2: BP > PC
Option 3: BP < PC
Option 4: BP = $\frac{1}{2}$ PC
Correct Answer: BP = PC
Solution :
We know, AB, BC, and AC are tangents to the circle.
Let AB, BC, and AC touch the circle at points Q, P, and R respectively.
So, AQ = AR = a (say) (length of tangents from the same point are equal)
And BQ = BP = b (say) (length of tangents from the same point are equal)
And CP = CR = c (say) (length of tangents from the same point are equal)
Now, AB = a + b and AC = a + c
Since AB = AC, a + b = b + c
Or, a = c
Hence, the correct answer is BP = PC.
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