Question

In the given figure, MNO is an isosceles triangle with MN = MO and MO is tangent to the circle at P. If AM = 6 cm and MP = PO, then the length of MO is

Answer 1

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$\bold{Explanation:}$

Given: Δ LMN is an isosceles triangle, m∠LMN = m∠LNM

And, LP bisects angle NLQ.

Prove: LP ║ MN

Now, In triangle LMN,

m∠LMN + m∠LNM+ m∠MLN = 180° (By the property of triangle)

⇒ m∠LMN + m∠LMN + m∠MLN = 180° ( Here,  m∠M = m∠N )

⇒ 2 m∠LMN + m∠MLN = 180° -------(1)

Now, LP bisects angle NLQ.

⇒ m∠PLN =m∠QLP ( by the property of angle bisector)

Since, m∠QLP + m∠PLN + m∠MLN = 180° ( sum of all angles on a straight line)

m∠QLP + m∠QLP + m∠MLN = 180°

2 m∠QLP + m∠MLN = 180°

⇒ m∠MLN = 180°- 2 m∠QLP --------(2)

From equation (1) and (2),

2 m∠LMN + 180°- 2 m∠QLP = 180°

2 m∠LMN - 2 m∠QLP = 0

2 m∠LMN = 2 m∠QLP

m∠LMN = m∠QLP

⇒ ∠LMN ≅ ∠QLP

Thus, By the inverse of corresponding angle theorem,

LP ║ MN

Answer 2

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