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

In a DeltaABCDeltaABC, right angled at AA, a point DD is on side ABAB. Prove that CD^2=BC^2+BD^2CD^2=BC^2+BD^2?