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can anyone please help me with this question??
ABC is a right angle triangle where angle A is right angle. The perpendicular
drawn from A on BC intersects BC at point D. A point P is chosen on the circle
drawn through the vertices of ΔADC such that CP perpendicular to BC and AP = AD. If a
square is drawn on the side BP, the area is 340 square units. What is the area of
triangle ABC?

The answer is 68. First u need to prove that the quadrilateral ADPC is a square. Then u will see that (BPC) = (ABC) Then using pythagorus derive BC. The result follows by deriving BC.

Note that $ADCP$ is a square and $CD = \frac{1}{2}CB = CP = DA$

$BP^2 = CB^2 + CP^2 = CD^2 + (2CD)^2 = 5CD^2 = 340 \Longrightarrow CD^2 = 68 \Longrightarrow CD = 2\sqrt{17}$

So, $\bigtriangleup ABC = \frac{1}{2} * CB * AD = CD * CD = 68$