Write another proof of the above theorem (property of an angle bisecto

1.	Write another proof of the above theorem (property of an angle bisector of a triangle). Use the following properties and write the proof. i. The areas of two triangles of equal height are proportional to their bases. ii. Every point on the bisector of an angle is equidistant from the sides of the angle.Given: In ∆CAB, ray AD bisects ∠A. To prove: = AB/AC = BD/DCConstruction: Draw seg DM ⊥ seg AB A – M – B and seg DN ⊥ seg AC, A – N – C.
Answer» Proof: In ∆ABC, Point D is on angle bisector of ∠A. [Given] ∴DM = DN [Every point on the bisector of an angle is equidistant from the sides of the angle] A(∆ABD)/A(∆ACD) = (AB x DM) (AC x DN) [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights] ∴ A(∆ABD)/A(∆ACD) = AB/AC (ii) [From (i)] Also, ∆ABD and ∆ACD have equal height. ∴ A(∆ABD)/A(∆ACD) = BD/CD (iii) [Triangles having equal height ] ∴ AB/AC = BD/DC [From (ii) and (iii)]

Write another proof of the above theorem (property of an angle bisector of a triangle). Use the following properties and write the proof. i. The areas of two triangles of equal height are proportional to their bases. ii. Every point on the bisector of an angle is equidistant from the sides of the angle.Given: In ∆CAB, ray AD bisects ∠A. To prove: = AB/AC = BD/DCConstruction: Draw seg DM ⊥ seg AB A – M – B and seg DN ⊥ seg AC, A – N – C.

Answer»

Proof:

In ∆ABC,

Point D is on angle bisector of ∠A. [Given]

∴DM = DN [Every point on the bisector of an angle is equidistant from the sides of the angle]

A(∆ABD)/A(∆ACD) = (AB x DM) (AC x DN) [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]

∴ A(∆ABD)/A(∆ACD) = AB/AC (ii) [From (i)]

Also, ∆ABD and ∆ACD have equal height.

∴ A(∆ABD)/A(∆ACD) = BD/CD (iii) [Triangles having equal height ]

∴ AB/AC = BD/DC [From (ii) and (iii)]