In triangle ABC, angle B =90° seg BD is perpendicular to side AC. A-D-C. If AD = 8 and DC= 2cm , find BD
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Let ∠ACB=x,⇒∠CBD=90−x
⇒∠ABD=x,⇒∠BAD=90−x
⇒ΔABC,ΔADB,andΔBDC are similar
⇒BDCD=ADDB
⇒8CD=48
⇒CD=8×84=16
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