Question

A AOB - 600 Cher LDAB = 40", ZADb-To
and LOBH=60', Thentina ZODC, 2006, 2000
C
B​

Answer 1

Answer:

Explanation:

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