Question

In triangle ABC,angle A =2 times angle B.Prove that BC2 =AC2 +AC.AB.

Answer 1

Given :-

• ∠A = 2 . ∠B

To Prove :-

• BC² =AC² +AC.AB

Formula used :-

if sides opposite to ∠A, ∠B, ∠C be a,b,c respectively.

Than,

By sine rule :- a/(Sin A)= b/(Sin B)= c/(Sin C)

and,

By cosine rule :-

• cos B = ( a² + c² - b² / 2ac )

Solution :-

From sin rule : -

→ BC/sin A = AC/Sin B

Given that, ∠A = 2 . ∠B

So,

→ BC/sin.2B = AC/Sin B

using , sin2@ = [email protected]@ we get,

→ BC / (2 * sin B * cos B) = AC/Sin B

→ BC / 2.cos B = AC

→ cos B = BC/2.AC -------------- Eqn.(1)

From Cosine rule :-

→ cos B = (AB² + BC² - AC²)/ (2.AB.BC) ------ Eqn.(2)

Comparing Both Equations Now, we get,

→ BC/2.AC = (AB² + BC² - AC²)/ (2.AB.BC)

→ 2.AB.BC² = 2.AC(AB² + BC² - AC²)

→ AB.BC² = AC.AB² + AC.BC² - AC³

→ AB.BC² - AC.BC² = AC.AB² - AC³

→ BC²(AB - AC) = AC(AB² - AC²)

using (x² - y²) = (x + y)(x - y) in RHS now,

→ BC²(AB - AC) = AC (AB + AC)(AB - AC)

→ BC² = AC(AB + AC)

→ BC² = AC.AB + AC²

→ BC² =AC² + AC.AB (Proved).

Answer 2

Answer:

Given :

• ∆ ABC
• Let us assume that,
• ∠A = 2x
• ∠B = x

To Prove :

BC ² = AC²+AB×AC

Construction:

CX is perpendicular to AB

and also construct,

CD = CA ...........(1)

Proof :

∠A = ∠D = 2X

In ∆CDB,

Exterior ∠2x = ∠B + ∠C

→∠B = ∠BCD = X

→CD = DB ........(2) [ By side opposite to equal angle of a triangle]

From (1) and (2)

AC = DB .........(3)

By Pythagoras theorem of acute triangle,

BC² = AC²+ AB²-2 AB×AX

BC²= AC²+AB (AB -2AX)

BC²= AC²+AB(AB-AD)

BC²= AC²+AB×DB

But in equation (3)

DB = AC

HENCE,

BC²= AC²+AB×AC

Hence Proved!!!