Theorem 6.9: In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle. Given: A triangle ABC in which
γπ΄πΆγ^2=γπ΄π΅γ^2+γπ΅πΆγ^2
To Prove: β B=90Β°
Construction: Draw Ξ PQR right angled at Q, such that
PQ = AB and QR = BC.
Proof:
In βPQR
β Q=90Β°
By Pythagoras theorem,
γππ γ^2=γππγ^2+γππ γ^2
Since PQ = AB and QR = BC
γππ γ^2=γπ΄π΅γ^2+γπ΅πΆγ^2
Also, given that
γπ΄πΆγ^2=γπ΄π΅γ^2+γπ΅πΆγ^2
From (1) & (2)
γππ γ^2=γπ΄πΆγ^2
PR = AC
In Ξ ABC & Ξ PQR
AC = PR
AB = PQ
BC = QR
β΄ Ξ ABC β Ξ PQR
β β B = β Q
Since β Q = 90Β°
β΄ β B = 90Β°
Hence Proved.

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.