In a #DeltaABC#, right angled at #A#, a point #D# is on side #AB#. Prove that #CD^2=BC^2+BD^2#?

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Answer 1 · 2016-12-06T12:51:31

#CD^2!=BC^2+BD^2#, but

#CD^2=BC^2+BD^2-2BDxxAB#

With a point #D# on #AB#, we have the figure as
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In the right angled triangle #ABC# right angled at #A#, we have

#BC^2=AB^2+AC^2=(AD+BD)^2+AC^2#

= #AD^2+BD^2+2xxBDxxAD+AC^2#

or #BC^2+BD^2#

= #AD^2+AC^2+2BD^2+2xxBDxxAD#

= #CD^2+2BD(BD+AD)#

= #CD^2+2BDxxAB#

or #CD^2=BC^2+BD^2-2BDxxAB#