# Median of Right triangle property

The property states that ” in right triangle, the length of median that touches hypotenuse is half of the length of hypotenuse “.

Consider the above right triangle ABC in which ∠ABC = 90 degree.

Here BM is the median of triangle which divide the hypotenuse into two equal parts, AM = MC.

According to the theorem, the length of median BM is half of hypotenuse AC.

i.e. \mathtt{BM=\frac{1}{2} .AC}

### Proof of median of right triangle property

Given:
Given above is right triangle ABC.
∠ABC = 90 degree

BM is the median intersecting the hypotenuse AC.
i.e. AM = MC.

Construction:
Draw line MN such that it is parallel to line BC.
i.e. MN || BC

To prove:
Median is half of hypotenuse.
\mathtt{BM=\frac{1}{2} .AC}

Proof:
Consider triangle ABC.

M is the midpoint of AC { given }
Line MN || BC { construction }

Using the midpoint theorem of triangle, we can say that point N is midpoint of AB.
i.e. AN = NB

We know that NM || BC.

Since, BC is perpendicular to AB. We can say that NM is also perpendicular to AB.

i.e. ∠ANM = 90 degree.

Take triangle ANM and BNM

AN = NB { Proved above }
NM = MN { common side }
∠ANM = ∠BNM = 90 degree

By SAS congruency condition we can say that both triangles are congruent.
Hence, \mathtt{\triangle ANM\cong \triangle BNM}

Since both triangles are congruent, we can say that;
BM = AM

This can also be written as; \mathtt{BM=\frac{1}{2} .AC}

Hence, we proved that in right triangle, the length of median of hypotenuse is half of half of hypotenuse.

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