In triangle ABC AB =AC , BD perpendicular AC and CF perpendicular AB prove that Triangle ABD congruent to Triangle ACF
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In Δ BCD by pythagoras theorem, we have
⇒ BC2 = BD2 + CD2 ... (1)
Again, in Δ ABD by pythagoras theorem
AB2 = BD2 + AD2
⇒ BD2 = AB2 - AD2
On putting value of BD2 in (1), we get
BC2 = AB2 - AD2 + CD2
⇒ BC2 = AB2 - (AC - CD)2 + CD2
⇒ BC2 = AB2 - AC2 - CD2 + 2AC. CD + CD2
⇒ BC2 = 2AC. CD [Given, AB = AC. ⇒ AB2 = AC2]
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