Question

If P is any interior point of ΔABC, prove that ∠BPC > ∠BAC.

Answer 1

Given:

• P is any interior point of triangle ABC.

To Prove:

• ∠BPC > ∠BAC.

Proof [Step-wise]:

1. Let the six angles at the vertices formed by P be:

∠PAB = a, ∠PAC = a’ (so a + a’ = ∠BAC),

∠PBA = b, ∠PBC = b’ (so b + b’ = ∠ABC),

∠PCB = c, ∠PCA = c’ (so c + c’ = ∠ACB).

2. In triangle PBC,

The angle at P is ∠BPC 

= 180° – (∠PBC + ∠PCB)

= 180° – (b’ + c)

3. Consider the sum of the six small angles around A, B, C:

a + a’ + b + b’ + c + c’

= ∠A + ∠B + ∠C

= 180°

Hence, (a + a’) + (b’ + c) + (c’ + b) = 180°.

4. Compute the difference:

∠BPC – ∠BAC

= [180° – (b’ + c)] – (a + a’) 

= 180° – [(a + a’) + (b’ + c)]

From step 3,

(a + a’) + (b’ + c)

= 180° – (b + c’) 

Therefore, ∠BPC – ∠BAC = b + c’.

5. Since P is an interior point, the angles b (∠PBA) and c’ (∠PCA) are positive.

Thus, b + c’ > 0.

So, ∠BPC – ∠BAC > 0.

i.e. ∠BPC > ∠BAC.

For any interior point P of triangle ABC, ∠BPC > ∠BAC, as required.