Question

Show that of all the line segments drawn from a given point not on it , the perpendicular line segment is the shortest.

Answer 1

Step-by-step explanation:

Let L is a line And A is a point not on the line L.

Now draw AC perpendicular to L.

Again let D is a point on L other than C.

Now in triangle ACD

∠C = 90

Since ∠A + ∠C + ∠D = 180

=> ∠A + ∠D = 180 - ∠C

=> ∠A + ∠D = 180 - 90

=> ∠A + ∠D = 90

So ∠D is an acute angle.

So ∠C > ∠D

=> AD > AC (since opposite to greater angle is greater)

=> AC < AD

So all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Answer 2

Answer:

Let L is a line And A is a point not on the line L.

Now draw AC perpendicular to L.

Again let D is a point on L other than C.

Now in triangle ACD

∠C = 90

Since ∠A + ∠C + ∠D = 180

=> ∠A + ∠D = 180 - ∠C

=> ∠A + ∠D = 180 - 90

=> ∠A + ∠D = 90

So ∠D is an acute angle.

So ∠C > ∠D

=> AD > AC (since opposite to greater angle is greater)

=> AC < AD

So all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.