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ABC is triangle. D is the middle point o...

ABC is triangle. D is the middle point of BC. If AD is perendicular to AC, then prove that
`cosAcosC=(2(c^(2)-a^(2)))/(3ac)`

Text Solution

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In `Delta ADC,` we have

`cos C=(AC)/(CD)`
`cos C=(2b)/(a)" "...(i)`
Applying cosine formula in `DeltaABC`, we have
`cosA =(b^(2)+c^(2)-a^(2))/(2bc)`
and `cos C=(a^(2)+b^(2)-c^(2))/(2ab)" "....(ii)`
From Eqs.(i) and (ii),
`(a^(2)+b^(2)-c^(2))/(2ab)=(2b)/(a)`
`rArr a^(2)+b^(2)-c^(2)=4b^(2)`
`rArr a^(2)-c^(2)=3b^(2)" "....(iii)`
Now, ` cos A cos C=(b^(2)+c^(2)-a^(2))/(2bc)*(2b)/(a)`
`(b^(2)+c^(2)-a^(2))/(bc)=(3b^(2)+3(c^(2)-a^(2)))/(3ac)`
`((a^(2)+c^(2))+3(c^(2)-a^(2)))/(3ac)=(2(c^(2)-a^(2)))/(3ac)`
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