Question : In $\triangle A B C$, M is the midpoint of the side AB. N is a point in the interior of $\triangle A B C$ such that CN is the bisector of $\angle C$ and $C N \perp N B$. What is the length (in cm) of MN, if BC = 10 cm and AC = 15 cm?
Option 1: 5
Option 2: 4
Option 3: 2.5
Option 4: 2
Correct Answer: 2.5
Solution :
In $\triangle ABC, AM = MB$
In $\triangle NPC$ and $\triangle NBC,$
⇒ $\angle BCN = \angle NCP$
⇒ $\angle CNB = \angle CNP = 90^{\circ}$
⇒ NC = NC (common side)
$\therefore \triangle NPC$ is congruent to $\triangle NBC$
NB = NP
BC = PC = 10 cm
AP = AC – PC = 15 – 10 = 5 cm
In $\triangle ABP$,
M and N are midpoints of AB and BP
By the midpoint theorem,
$MN = \frac{AP}{2} = \frac{5}{2} = 2.5$ cm
Hence, the correct answer is 2.5.
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