Prove that the areas of two similar triangles are in the ratio of the squares of the
corresponding angle bisectors.
the lines
Answers
Answer:
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Answer:
ANSWER
If two triangles are similar, then the ratio of the area of both triangles is proportional to square of the ratio of their corresponding sides.
To prove this theorem, consider two similar triangles ΔABC and ΔPQR
According to the stated theorem
ar△PQR
ar△ABC
=(
PQ
AB
)
2
=(
QR
BC
)
2
=(
RP
CA
)
2
Since area of triangle =
2
1
×base×altitude
To find the area of ΔABC and ΔPQR draw the altitudes AD and PE from the vertex A and P of ΔABC and ΔPQR
Now, area of ΔABC =
2
1
×BC×AD
area of ΔPQR =
2
1
×QR×PE
The ratio of the areas of both the triangles can now be given as:
ar△PQR
ar△ABC
=
2
1
×QR×PE
2
1
×BC×AD
ar△PQR
ar△ABC
=
QR×PE
BC×AD
Now in △ABD and △PQE it can be seen
∠ABC=∠PQR (Since ΔABC∼ΔPQR )
∠ADB=∠PEQ (Since both the angles are 90°)
From AA criterion of similarity ΔADB∼ΔPEQ
PE
AD
=
PQ
AB
Since it is known that ΔABC∼ΔPQR
PQ
AB
=
QR
BC
=
RP
CA
Substituting this value in equation , we get
ar△PQR
ar△ABC
=
PQ
AB
×
PE
AD
we can write
ar△PQR
ar△ABC
=(
PQ
AB
)
2
Similarly we can prove
ar△PQR
ar△ABC
=(
PQ
AB
)
2
=(
QR
BC
)
2
=(
RP
CA
)
2
Step-by-step explanation:
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