What is 'APPOLONIUS THEOREM'?
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In geometry, Apollonius's theorem is a theorem relating the length of a median of a triangle to the lengths of its side. It states that "the sum of the squares of any two sides of any triangle equals twice the square on half the third side, together with twice the square on the median bisecting the third side".
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Heya !!
Here's your answer..⬇⬇
_______________________________
⚫ Apolloneous Theorem :-
Apolloneous theorem related with length of Median of a triangle to length of its sides.
⚫ To Prove Apolloneous theorem :-
Let in ∆ABC, AM is a Median ( BM = MC ) and AH is perpendicular to the Base BC.
Apolloneous theorem :-
AB^2 + AC^2 = 2( AM^2 + BM^2 )
⚫ Proof :- In ∆AHB, angle AHB = 90°
AB^2 = AH^2 + HB^2 ---- ( 1 )
In ∆AHC, angle AHC = 90°
AC^2 = AH^2 + HC^2 ----- ( 2 )
Adding L.H.S ..,
AB^2 + AC^2 = AH^2 + HB^2 + AH^2 + HC^2
= 2AH^2 + BH^2 + HC^2
= 2 ( AM^2 - MH^2 ) + ( BM - HM )^2 + ( CM + HM )^2
= 2AM^2 - 2MH^2 + BM^2 - 2BM*HM + HM^2 + CM^2 + 2CM*HM
= 2AM^2 + BM^2 + CM^2
= 2AM^2 + 2BM^2 -- ( BM = CM )
= 2( AM^2 + BM^2 )
Hence, proved that AB^2 + AC^2 = 2( AM^2 + BM^2 ).
________________________________
Hope it helps..
Thanks :))
Here's your answer..⬇⬇
_______________________________
⚫ Apolloneous Theorem :-
Apolloneous theorem related with length of Median of a triangle to length of its sides.
⚫ To Prove Apolloneous theorem :-
Let in ∆ABC, AM is a Median ( BM = MC ) and AH is perpendicular to the Base BC.
Apolloneous theorem :-
AB^2 + AC^2 = 2( AM^2 + BM^2 )
⚫ Proof :- In ∆AHB, angle AHB = 90°
AB^2 = AH^2 + HB^2 ---- ( 1 )
In ∆AHC, angle AHC = 90°
AC^2 = AH^2 + HC^2 ----- ( 2 )
Adding L.H.S ..,
AB^2 + AC^2 = AH^2 + HB^2 + AH^2 + HC^2
= 2AH^2 + BH^2 + HC^2
= 2 ( AM^2 - MH^2 ) + ( BM - HM )^2 + ( CM + HM )^2
= 2AM^2 - 2MH^2 + BM^2 - 2BM*HM + HM^2 + CM^2 + 2CM*HM
= 2AM^2 + BM^2 + CM^2
= 2AM^2 + 2BM^2 -- ( BM = CM )
= 2( AM^2 + BM^2 )
Hence, proved that AB^2 + AC^2 = 2( AM^2 + BM^2 ).
________________________________
Hope it helps..
Thanks :))
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