In an isosceles trapezoid ABCD, the extensions of the lateral sides AB and CD intersect at point P. Find the height PK

In an isosceles trapezoid ABCD, the extensions of the lateral sides AB and CD intersect at point P. Find the height PK of triangle PBC if AB = 13, AD = 18, and the height of the trapezoid BH = 12.

From the right-angled triangle ABN, we determine the length of the leg AH.

AH ^ 2 = AB ^ 2 – BH ^ 2 = 169 – 144 = 25.

AH = 5 cm.

The height of the PM divides the base AD in half, then AM = AD / 2 = 18/2 = 9 cm.

Let us prove that right-angled triangles ABH and APM are similar.

Angle A is common for triangles, then triangles are similar in the first sign of similarity to right-angled triangles – in an acute angle.

Then BH / PM = AH / AM.

12 / PK = 5/9.

PM = 9 * 12/5 = 21.6 cm.

Then PK = PM – MK = 21.6 – 12 = 9.6 cm.

Answer: The height PK is 9.6 cm.