Question

Prove phythagoras theorem

Answer 1

consider a right angle triangle ABC , right angled at B .
in triangle ABC ,
SINθ = AB / AC
COSθ = BC / AC
by squaring and adding both sides
SIN^2 Θ + COS^2 Θ = (AB / AC) ^2 + (BC /AC)^2
1 = (AB^2 + BC^2) / AC^2
AC^2 = AB^2 + BC^2
as AC is H , AB is P & BC is B of triangle ABC
therefore H^2 = P^2 + B^2

Answer 2

Kindly refer the attachment which I had inserted above.

Assurance: 100% Reliable and of my own and not copied from other sources

Statement:
In a right angled triangle the square of hypotenuse is equal to the sum of the squares of the other two sides.

Given:
In a right angled triangle, ABC, A is the right angle.

To prove:
BC×BC = AB×AB + AC×AC

Construction:
Draw AD perpendicular to BC.

Proof:
In right triangle ABC and DBA, B is the common angle. And also we have angle A and angle D is right angle.

By AA similarity Criterion,
∆ABC ~ ∆DBA

As their sides are proportional,

Hence, AB/DB = BC/BA
AB × AB = DB × BC. ====> 1

Similarly, we have ∆ABC ~ ∆DAC,

Thus,. BC/AC = AC/DC

AC×AC = BC × DC. =====> 2

By adding 1 and 2,

(AB×AB) + (AC× AC) = (BD×BC) + (BC×DC)

= BC ( BD + DC )

= BC ( BC)

Thus, BC×BC = AB×AB + AC×AC.

Hence the Pythagoras theorem.