Question

In triangle ABC , AD bisects angle A and Angle B is twice of angle C . prove that angle A equal to 72° . the Constructions are made in the figure already to help you. Best answer will receive extra 20 points , wrong ans will be punished

Answer 1

Answer:

From the figure, In ∆ ABC, we have

Angle CAD = angle DAB = ½ * angle BAC= ½ * 2y = y [∵ AD bisects angle A]

 

Angle B is given to be twice of angle C, so

If Angle B = 2x then angle C = x

From the figure, we can consider BE bisect angle B, therefore,  

Angle ABE = angle EBC = ½ * angle ABC = ½ * 2x = x

In ∆BCE,  

Angle EBC = angle ECB = x

∴ CE = BE …… (i) [∵ sides opposite to equal angles are equal]

Now, consider ∆ ABE and ∆DCE,  

AB = CD ….. [given in the figure]

Angle ABE = angle ECD = x  

CE = BE [from (i)]

∴ By SAS congruence criterion, we have

∆ ABE ≅ ∆DCE

And, by Corresponding Parts of Congruent Triangles i.e., CPCT, we get

Angle BAE = angle CDE = 2y  and AE = DE ….. (ii)

From (ii), ∵ AE = DE

∴ angle EAD = angle EDA = y ….. (iii)

In ∆ ABD, we have  

Angle ADC = angle DAB + angle ABD  .... [exterior angle property]

⇒ angle EDA + angle CDE = angle DAB + angle ABD

⇒ y + 2y = y + 2x ……  [from the values given in the figure and eq. (ii) & (iii)]

⇒ 3y = y + 2x  

⇒ 2y = 2x ……. (iv)

Thus, in ∆ ABC,

Angle A + angle B + angle C = 180° [∵ angle sum property of a triangle]

⇒ 2y + 2x + x = 180°

⇒ 2y + 2y + y = 180° ….. [from (iv)]

⇒ 5y = 180°

⇒ y = 36°

Thus, angle A = 2y = 2 * 36° = 72°

Hence proved  

Answer 2

Answer:

the angel 2y = 2 * 36° = 72°

Step-by-step explanation:

hope it was help for you