Question

In the adjoining figure, AD bisects Angle A. Arrange AB, BD and DC in the descending order of their lengths.​

Answer 1

Answer:

Given; △ABC, AD bisects ∠A,

In △ABC,

Sum of angles = 180

∠A+∠B+∠C=180

∠A+60+40=180

∠A=80

∘

∠BAD=∠DAC=40

∘

∠A=80

∘

, ∠C=40

∘

Since, ∠A>∠C

BC>AB (Sides opposite greater angles is greater) (1)

In △ADC

∠ACD=∠DAC=40

∘

Thus, AD=DC (Isosceles triangle property)

Now, In △ABD

Sum of angles = 180

∠ABD+∠ADB+∠BAD=180

60+∠ADB+40=180

∠ADB=80

∘

∠ABD=60

∘

and △ADB=80

∘

Since, ∠ABD>∠ADB

Thus, AD>AB

or DC>AB (Since, AD=DC) (2)

and we know BC=BD+DC

Hence, BC>DC (3)

Hence, from (1), (2) and (3)

BC>DC>AB