InFig.D is the midpoint of side BC and AE⊥BC.IfBC=aAC=b, AB=c,ED=x,AD= p and AE = h, prove that: (b2 – c2) = 2ax
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(i) In right-angled triangle AEC, applying Pythagoras theorem, we have:
(ii) In right-angled triangle AEB, applying Pythagoras, we have:
(iii) Adding (i) and (ii), we get:
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