Prove that, when two triangles are
similar ratio of areas of those
triangles is equal to the ratio of
the squares of their corresponding
i] medians
ii] altitude
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Answered by
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ANSWER
Given:
△ABC∼△DEF
O is a median of BC and P is a median of EF
To Prove:
A(△DEF)
A(△ABC)
=
(DP)
2
(AO)
2
Proof:
Since, △ABC∼△DEF
∴ ∠A=∠D, ∠B=∠E, ∠C=∠F (Corresponding Angles of Similar Triangles) ....(1)
Also,
DE
AB
=
EF
BC
=
DF
AC
(Corresponding Sides of Similar Triangles) ......(2)
Since, BC=2BO and EF=2EP
∴ Equation (2) can be written as,
DE
AB
=
EF
BC
=
DF
AC
=
EP
BO
......(3)
In △AOB and △DPE
∠B=∠E (From 1)
DE
AB
=
EP
BO
(From 3)
∴ By SAS Criterion of Similarity, △AOB ∼△DPE
∴
DE
AB
=
EF
BC
=
DF
AC
=
DP
AO
=Ratio of their heights ....(4) (Corresponding Sides of Similar Triangles)
A(△DEF)
A(△ABC)
=
2
1
×EF×Height
2
1
×BC×Height
=
(DP)
2
(AO)
2
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