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Question
In ΔABC, AB = AC. Side BC is produced to D. Prove that `AD^2−AC^2`= BD.CD
Solution
Draw AE⊥BC, meeting BC at D.
Applying Pythagoras theorem in right-angled triangle AED, we get:
Since, ABC is an isosceles triangle and AE is the altitude and we know that the altitude is also the median of the isosceles triangle.
So, BE = CE
And DE+CE=DE+BE=BD
`AD^2=AE^2+DE^2`
`⇒ AE^2=AD^2-DE^2 ` ...............(1)
In ΔACE,
`AC^2=AE^2+EC^2`
⇒ `AE^2=AC^2-EC^2 ` ...............(2)
Using (i) and (ii),
⇒` AD^2-DE^2=AC^2-EC^2`
⇒` AD^2-AC^2=DE^2-EC^2`
`=(DE+CE) (DE-CE)`
`(DE+BE) CD`
`BD.CD`
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