Problem:
Let △ABC be an isosceles triangle with ∠A=90∘. There exists a point P inside △ABC such that ∠PAB=∠PBC=∠PCA and AP=10. Find the area of △ABC.
Solution:
Answer (250):
Because ∠PCA and ∠PAB are both the complement of ∠PAC, it follows that ∠APC=90∘. Because △ABC is isosceles, ∠ABP=∠ABC−∠PBC=∠BCA−∠PCA=∠BCP, so △BPC∼△APB.
Because BC=2⋅AB, the similarity ratio is 2. Thus
PC=2⋅PB=2⋅2⋅PA=2⋅PA=20
and the area of △ABC equals
21AC2=21(PA2+PC2)=21(102+202)=250.
OR
As above, ∠APC=90∘ and △BCP∼△ABP. Thus ∠APB=∠BPC=21(360∘−∠APC)=135∘. It follows that the diagram fits into a 20×30 grid, as shown, from which the area can be calculated as
20⋅30−21(10⋅20+10⋅20+10⋅30)=250
Note: The point P is known as the first Brocard point for △ABC. Angle ∠PAB is approximately 26.565∘.
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