Prove Parallelogram ABCD: Triangle APD = ABP + DCP

[jtf2eh]

Parallelogram proof :(

Homework Statement

In parallelogram ABCD, P is any point on BC. Prove that triangle APD = triangle ABP + triangle DCP

Homework Equations

n/a

The Attempt at a Solution

I really don't know where to start :(

any help would be greatly appreciated :)

 

[HallsofIvy]

First do you mean that the areas satisfy that equation? If so just note that the lengths of the bases of the two triangles ABP and DCP add to the length of the base of triangle APD and all three triangles have the same height.

 

[Architeuthis Dux]

To prove that triangle APD is equal to the sum of triangle ABP and DCP, we can use the fact that the opposite sides of a parallelogram are equal in length and parallel to each other.

First, let's draw a diagram of parallelogram ABCD with point P on BC.

Now, we can draw a line segment from point P to point A, creating triangle APD.

Next, we can draw a line segment from point P to point B, creating triangle ABP.

And finally, we can draw a line segment from point P to point C, creating triangle DCP.

Now, we can see that triangle APD shares one side with both triangle ABP and DCP, which is side AP.

Using the fact that opposite sides of a parallelogram are equal in length, we can say that side AP is equal in length to side AD and side CP.

Therefore, we can rewrite triangle APD as triangle ADP and rewrite triangle DCP as triangle PCD.

Now, we can see that triangle ADP is equal to the sum of triangle ABP and triangle PCD, as they share two sides, AD and AP.

And since triangle ADP is equal to triangle APD, we can replace triangle APD in the previous statement, giving us:

Triangle APD = triangle ABP + triangle DCP

Thus, we have proven that triangle APD is equal to the sum of triangle ABP and DCP in parallelogram ABCD.