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.