Let ABCD be a rectangle P is placed in the interior of the r

Let ABCD be a rectangle. P is placed in the interior of the rectangle. P is connected to A and D and to points B and C. The ratio of the sum of the area of the triangles APD and BPC and the area of ABCD is found to be 1/2 no matter where P is moved within ABCD. I need to prove why the ratio is always 1/2.

Solution

Since the given figure is a rectangle, hence the opposite sides will be equal

Therefore we can write

AD = BC and AB=CD

Area of Rectangle ABCD = AB * BC

Let us drop the altitude from point P to the base AD and base BC as well

Let us assume that altitude from P to the base AD intersects it at h and base BC is intersected at h\'

Area of triangle ADP = 1/2 * Ph * AD

Area of triangle BPC = 1/2 * Ph\' * BC

Area of triangle ADP + Area of triangle BPC = 1/2 * Ph * AD + 1/2 * Ph\' * BC

since ABCD is a rectangle, hence AD = BC and even Ph\' + Ph = CD, as shown in the figure

=> 1/2 * (Ph+Ph\') * BC

=> 1/2 * (AB) * BC

=> 1/2 * (area of rectangle)