In ΔABC, M is the midpoint of the side AB. N is a point in the interior of ΔABC such that CN is the bisector of ∠C and CN ⊥ NB. What is the length (in cm) of MN, if BC = 10 cm and AC = 15 cm?

Given:

In ΔABC, M is the midpoint of the side AB

N is a point in the interior of ΔABC such that CN is the bisector of ∠C and CN ⊥ NB

BC = 10 cm

AC = 15 cm

Concept used:

Midpoint theorem - The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side

Calculation:

Construction: Produce BN to P which meets AC at P.

And Join MN

According to the question

In ΔNPC and ΔNBC

∠N = ∠N  [90°]

BC = PC [corresponding side]

BN = NP [corresponding angle]

⇒ ΔNPC ≅ ΔNBC 

Hence, NB = NP (It means Point N is the midpoint of side BP)

And BC = PC = 10 cm

So, AP = AC – PC

⇒ AP = (15 – 10) cm

⇒ AP = 5 cm

Now, In ΔABP

M and N are the midpoints of AB and BP

So, According to the midpoint theorem

⇒ MN = \(\frac{AP}{{2}}\)

⇒ \(\frac{5}{{2}}\) cm

⇒ 2.5 cm

∴ The length of MN is 2.5 cm

Shortcut Trick  

The using mid-point theorem,

In ΔBAP

MN = \(AP\over2\) = \(\frac{5}{{2}}\) = 2.5 cm