Given : In ΔABC,AB = 20 cm, AC = 15 cm and the length of the altitude AN is 12 cm.
To prove: Triangle ABC is a right angled triangle
Proof:
Since AN is an altitude, so it is a line segment through a vertex and perpendicular to a line containing the base.
Consider the triangle ANB,
By Pythagoras theorem,
[tex] (AB)^2 = (AN)^2+(BN)^2 [/tex]
[tex] (20)^2 = (12)^2+(BN)^2 [/tex]
[tex] (BN)^2=400-144 [/tex]
[tex] (BN)^2 = 256 [/tex]
So, BN = 16 cm.
Consider the triangle ANC,
By Pythagoras theorem,
[tex] (AC)^2 = (AN)^2+(NC)^2 [/tex]
[tex] (15)^2 = (12)^2+(NC)^2 [/tex]
[tex] (NC)^2=225-144 [/tex]
[tex] (NC)^2 = 81 [/tex]
So, NC = 9 cm.
Therefore, BC = BN + NC = 16 + 9 = 25cm.
Now, consider [tex] (AB)^2 + (AC)^2=(BC)^2 [/tex]
[tex] (20)^2 + (15)^2=(25)^2 [/tex]
225 = 225
Therefore, By converse of Pythagoras theorem.
Triangle ABC is a right angled triangle.
