Question

proof phythagorus of 10th class ​

Answer 1

Draw a triangle ABC right angled at B.

For this we drop a perpendicular BD onto the side AC

We know, △ADB∼△ABC

Therefore, ADAB=ABAC (Condition for similarity)

Or, AB^2=AD×AC……..(1)

Also, △BDC∼△ABC

Therefore, CDBC=BCAC (Condition for similarity)

Or, BC^2=CD×AC……..(2)

Adding the equations (1) and (2) we get,

AB^2+BC^2=AD×AC+CD×AC

AB^2+BC2=AC(AD+CD)

Since, AD + CD = AC

Therefore, AC^2=AB^2+BC^2