Question

If the bisector of an angle of a triangle also bisects the opposite side, prove that the triangle is isosceles...​

Answer 1

Consider the ∆ABC, let AD be the bisector of ∠A and BD = CD. It is required to prove ∆ABC is an isosceles triangle i.e. AB = AC. For this draw a line from C parallel AD and extend BA. Let they meet at E.

It is given that

∠BAD = ∠CAD ... (1)

CE||AD

∴ ∠BAD = ∠AEC (Corresponding angles) ... (2)

And ∠CAD = ∠ACE (Alternate interior angles) ... (3)

From (1), (2) and (3)

∠ACE = ∠AEC

In ∆ACE, ∠ACE = ∠AEC

∴ AE = OAC (Sides opposite to equal angles) ... (4)

In ∆BEC, AD||CE and D is the mid-point of BC using converse of mid-point theorem A is the mid-point of BE.

∴ AB = AE

⇒ AB = AC [Using (4)]

In ∆ABC, AB = AC

∴ ∆ABC is an isosceles triangle.

Hence, proved

Answer 2

Answer:

Given, △ABC, AD bisects ∠A and AD bisects BC.

To prove: ABC is an Isosceles triangle

In △ABD and △ACD

∠DAB=∠DAC (AD bisects ∠A)

AD=AD (Common)

BD=CD (AD bisects BC)

Thus, △ABD≅△ACD (SAS rule)

Thus, AB=AC (By cpct)

hence, ABC is an Isosceles triangle.

Step-by-step explanation:

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