proof phythagorus of 10th class
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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
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