Question 33

If we join a vertex to a point on opposite side which divides that side in the ratio 1 : 1, then what is the special name of that line segment?

a) Median
b) Angle bisector
c) Altitude
d) Hypotenuse

Open in App

Solution

Consider $Δ A B C$ in which AD divides BC in the ratio 1 : 1

Now, BD : DC = 1 : 1
$\Rightarrow \frac{B D}{D C} = \frac{1}{1}$
$∴$ BD = DC
Since, AD divides BC into two equal parts. Hence, AD is the median.

Suggest Corrections

Similar questions

Q. Question 14
In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.

Q. In a right angled triangle, prove that the line segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.

If the line through one vertex of the triangle internally divides the opposite side such that they are proportional to the sides containing the angle, then the line must bisect the vertex angle.

What is the ratio that relates the opposite side of an angle to the hypotenuse in a right-angled triangle?

Q. An altitude has one end point at a vertex of the triangle and the other on the side which is opposite to that vertex , then how many altitudes does a triangle have?

Join BYJU'S Learning Program

Explore more

Median

Standard VII Mathematics

Join BYJU'S Learning Program

Question 33 If we join a vertex to a point on opposite side which divides that side in the ratio 1 : 1, then what is the special name of that line segment? a) Median b) Angle bisector c) Altitude d) Hypotenuse

Consider ΔABC in which AD divides BC in the ratio 1 : 1 Now, BD : DC = 1 : 1 ⇒BDDC=11 ∴ BD = DC Since, AD divides BC into two equal parts. Hence, AD is the median.

Question 33

If we join a vertex to a point on opposite side which divides that side in the ratio 1 : 1, then what is the special name of that line segment?

a) Median
b) Angle bisector
c) Altitude
d) Hypotenuse

Consider $Δ A B C$ in which AD divides BC in the ratio 1 : 1

Now, BD : DC = 1 : 1
$\Rightarrow \frac{B D}{D C} = \frac{1}{1}$
$∴$ BD = DC
Since, AD divides BC into two equal parts. Hence, AD is the median.