In trapezoid ABCD, point M is the midpoint of the larger base AD, point P is the midpoint of the lateral side AB.

In trapezoid ABCD, point M is the midpoint of the larger base AD, point P is the midpoint of the lateral side AB. Lines CM and PD meet at point K. Prove that the area of the quadrilateral APKM is equal to the area of the triangle KCD.

Since point P is the middle of the segment AB, you can draw the middle line of the trapezoid PP1.

From point P we draw a perpendicular PH to the base of the AD trapezoid, and from point C, a perpendicular CC1.

By the property of the middle line of the trapezoid, it divides in half any segment enclosed between the bases of the trapezoid, therefore CO = C1O = CC1 / 2.

Since PH is parallel to OC1, as perpendiculars to AD, and RO is parallel to HC1, then PHOC1 is a rectangle and PH = OC1 = OC = CC1 / 2.

Point M, by condition, is the middle of AD, therefore, AM = MD = AD / 2.

Determine the area of ​​the triangle APD.

Sard = AD * PH / 2.

Determine the area of ​​the triangle APD.

Ssmd = MD * CC1 / 2.

Let’s substitute in this equality the values ​​МD = АD / 2, and CC1 = 2 * PH.

Ssmd = (AD / 2) * 2 * RN / 2 = AD * RN / 2.

It turns out that Sard = Ssmd = AD * PH / 2.

The area of ​​both triangles includes the area of ​​the KMD triangle, then the area of ​​the ARKM quadrangle is equal to:

Sarkm = Sard – Skmd.

The area of ​​the triangle KCD is equal to:

Sksd = Ssmd – Skmd.

Since we have proved that Sard = Scmd, then Sksd = Sard – Scmd, and therefore is equal to Sarm.

Sarkm = Sksd, as required.