Question

show that the angle bisectors of a triangle are concurrent pls explain clearly​

Answer 1

Explanation:

Consider a triangle ABC. Let the angular bisectors of angles A & B meet at O.

From O, draw perpendiculars OD (to BC), OE (to AC) and OF (to AB)

Consider triangles BOD and BOF. Both are right angled triangles with the same hypotenuse, and with angle OBD = OBF (bisected angles of B) The two triangles are thus congruent and thus OD = OF.

By similar logic, by considering triangles AOF and AOE, it is possible to prove that OE=OF.

That is OD = OE = OF.

Now join OC.Consider the triangles OCD and OCE. Here again, both are right angled triangles with the same hypotenuse and (since OD=OE=OF) one more side also equal. Thus the two triangles are congruent and consequently, angle OCD=OCE.

Thus, a line joining the point -( of intersection of the bisectors of two of the angles in a triangle) - to the third angle, automatically bisects the third angle. Conversely, the angular bisectors of a triangle meet at a point or are concurrent.

Many thanks for the upvote.

Thanks to Kumar Saurav for his comments “Or, you could shorten the proof further, by using the fact that any point on the angle bisector of two lines is equidistant from the lines.” which I entirely concur with.

Answer 2

Answer:

Explanation:

Consider a triangle ABC. Let the angular bisectors of angles A & B meet at O.

From O, draw perpendiculars OD (to BC), OE (to AC) and OF (to AB)

Consider triangles BOD and BOF. Both are right angled triangles with the same hypotenuse, and with angle OBD = OBF (bisected angles of B) The two triangles are thus congruent and thus OD = OF.

By similar logic, by considering triangles AOF and AOE, it is possible to prove that OE=OF.

That is OD = OE = OF.

Now join OC.Consider the triangles OCD and OCE. Here again, both are right angled triangles with the same hypotenuse and (since OD=OE=OF) one more side also equal. Thus the two triangles are congruent and consequently, angle OCD=OCE.

Thus, a line joining the point -( of intersection of the bisectors of two of the angles in a triangle) - to the third angle, automatically bisects the third angle. Conversely, the angular bisectors of a triangle meet at a point or are concurrent.

Many thanks for the upvote.

Thanks to Kumar Saurav for his comments “Or, you could shorten the proof further, by using the fact that any point on the angle bisector of two lines is equidistant from the lines.” which I entirely concur with.