The perpendicular ad on the base bc of triangle abc intersects bc at d so that db=3cd. prove that 2ab^2=2ac^2+bc^2
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Draw the rough figure on your copy so that you will understand what am I writing
In triangle ABD
AB^2=AD^2+BD^2
but we have BD = 3/4BC DC=1/4BC
&AD^2=AC^2-DC^2=AC^2-(BC/4)^2
therefore AB^2=AC^2-(BC/4)+ (3BC/4)^2
AB^2=AC^2+BC^2/2
2AB^2=2AC^2+BC^2
In triangle ABD
AB^2=AD^2+BD^2
but we have BD = 3/4BC DC=1/4BC
&AD^2=AC^2-DC^2=AC^2-(BC/4)^2
therefore AB^2=AC^2-(BC/4)+ (3BC/4)^2
AB^2=AC^2+BC^2/2
2AB^2=2AC^2+BC^2
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