Question

triangle abc is isosceles in which AB = AC and d is a point on AC such that BC^2 = ACxCD . Prove that BD=BC

Answer 1

Answer:

Given :

In  Δ A B C

A B= A C  and  D  is a point on  A C  such that  B C  ×  B C  =  A C  × A D  

We have to prove

B D = B C  

Proof

Rearranging the given relation  

B C × B C = A C × A D

We can write

B C C D = A C B C

→ Δ A B C  ≅  Δ B D C  

Their corresponding angle pairs are:

1 . ∠ B A C = corresponding  ∠ D B C

2 . ∠ A B C = corresponding   ∠ B D C  

3 . ∠ A C B =corresponding   ∠ D C B

So as per above relation 2 we have  

∠ A B C =  corresponding  ∠ B D C  

Again in

Δ A B C

A B = A C

→ ∠ A B C = ∠ A C B = ∠ D C B

∴ In  Δ B D C ,

∠ B D C = ∠ B C D

→ B D = B C

Answer 2

Question:

→ ΔABC is isosceles in which AB = AC and D is a point on AC, such that BC² = AC x CD . Prove that BD = BC .

Step-by-step explanation:

Given:-

→ A △ABC in which AB = AC.

→ D is a point on AC such that BC² = AC × CD.

To prove :-

→ BD = BC  .

Proof :-

Since,

∵ BC² = AC × CD .

⇒ BC × BC = AC × CD  .

∴ AC/BC = BC/CD .......( i )  .

Also, ∠ACB = ∠BCD  . [ ∵ AB = AC , angles opposite to equal sides are equal . ]

∴ △ABC ~ △BDC.  [By SAS similarity rule. ]

∵ AB/AC = BD/BC

But, AB = AC ( Given ) .

⇒ AB/AB = BD/BC .

⇒ 1/1 = BD/BC .

∴ BD = BC.

Hence, it is proved .

THANKS .