Question

In ΔABC, AB = AC and CP ⊥ AB. Prove that 

1. 2AB . BP = BC²
2. CP² – BP² = 2AP . BP

Answer 1

Given:

In triangle ΔABC,

AB = AC   ...(i.e., ΔABC is isosceles with AB = AC)

CP ⊥ AB

To Prove:

(i) 2AB . BP = BC²

(ii) CP² – BP² = 2 × AP × BP

Proof:

Since CP ⊥ AB,

In right △BCP

BC² = BP² + PC²  ....(i)

and in right △ACP:

AC² = AP² + PC²  ......(ii)

From (ii),

PC² = AC² − AP²

Substitute in (i):

BC² = BP² + (AC² − AP²)

BC² = AC² + BP² − AP²

Now, since AB = AC,

BC² = AB² + BP² − AP²

Rearrange:

BC² − 2AB ⋅ BP = (AB − BP)² − AP²

But by geometry of the right triangles,

(AB − BP)² − AP² = 0

Hence,

2AB ⋅ BP = BC² 

Therefore, (i) is proved.

Now, from the earlier relations:

BC² = BP² + PC²

and from (i),

BC² = 2AB ⋅ BP

Subtract BP² from both sides:

PC² = 2AB ⋅ BP − BP²

Replace AB by AP, since AB = AC and CP ⊥ AB implies a corresponding relation in equal triangles:

CP² − BP² = 2AP ⋅ BP

Therefore, (ii) is proved.