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If perpendiculars from any point with an angle onits arms are congruent, prove that it lies on the bisector of that angle. |
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Answer» To prove-`/_1=/_2` Proof-`In /_PMB and /_PMB` PM=PN(given) BP=BD(common) `/_PMB=/_PNB(each 90^@)` `/_PMB cong /_PNB(RHS)` `/_1=/_2` BP bisects `/_B`. |
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