In ΔABC,AB = 20 cm, AC = 15 cm. The length of the altitude AN is 12 cm. Prove that ΔABC is a right triangle.

Refer to the attached image.Since AN is an altitude, an altitude of a triangle is a line segment through a vertex and perpendicular to a line containing the base forming a right angle with the base. Consider [tex] \Delta ABN [/tex],by Pythagoras theorem, we get[tex] (Hypotenuse)^{2}=(Base)^{2}+(Perpendicular)^{2} [/tex][tex] (AB)^{2}=(BN)^{2}+(AN)^{2} [/tex][tex] (20)^{2}=(BN)^{2}+(12)^{2} [/tex][tex] 400=(BN)^{2}+144 [/tex][tex] 400-144=(BN)^{2} [/tex][tex] (BN)^{2}=256 [/tex]So, BN = 16Consider [tex] \Delta ANC [/tex],by Pythagoras theorem, we get[tex] (Hypotenuse)^{2}=(Base)^{2}+(Perpendicular)^{2} [/tex][tex] (AC)^{2}=(NC)^{2}+(AN)^{2} [/tex][tex] (15)^{2}=(NC)^{2}+(12)^{2} [/tex][tex] 225=(NC)^{2}+144 [/tex][tex] 225-144=(NC)^{2} [/tex][tex] (NC)^{2}=81 [/tex]So, NC = 9So, BC = BN + NCBC = 16+9 = 25Now consider triangle ABC,Consider [tex] (BC)^{2}=(AB)^{2}+(AC)^{2} [/tex][tex] (25)^{2}=(20)^{2}+(15)^{2} [/tex]625 = 400 + 225625 = 625Therefore, by the converse of Pythagoras theorem , which states that "If the square of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle".Therefore, triangle ABC is a right triangle.