#### theorem

Theorem1: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given: ∆ABC ∼ ∆PQR

Construction: Draw AM ⊥ BC and PN ⊥ QR

To prove: `"ar∆ABC"/"ar∆PQR" = ("AB"/"PQ")^2= ("BC"/"QR")^2= ("AC"/"PR")^2`

Proof: ∆ABC ∼ ∆PQR (Given)

`"AB"/"PQ"= "BC"/"QR" = "AC"/"PR"` (Corresponding parts of Congruent triangles) ...eq1

and ∠B= ∠Q (Corresponding parts of Congruent triangles)

Now, `"ar(∆ABC)"= 1/2 × "BC" × "AM"`

`"ar(∆PQR)"= 1/2 × "QR" × "PN"`

`"ar(∆ABC)"/"ar(∆PQR)"= (1/2 xx "BC" xx "AM")/(1/2 xx "QR" xx "PN")`

`"ar(∆ABC)"/"ar(∆PQR)"`= `"BC"/"QR"×"AM"/"PN"` .......eq2

Again, In ∆ABM and ∆PQN

∠AMB = ∠PNQ = 90°

∠B= ∠Q (Corresponding parts of Congruent triangles)

∆ABM ∼ ∆PQN

So, `"AB"/"PQ"= "AM"/"PN"` (Corresponding parts of Congruent triangles) .....eq3

From eq2 and eq3

`"ar(∆ABC)"/"ar(∆PQR)"= "BC"/"QR"×"AB"/"PQ"` ......eq4

From eq1 and eq4

`"ar(∆ABC)"/"ar(∆PQR)"= "BC"/"QR" × "BC"/"QR"`

`"ar(∆ABC)"/"ar(∆PQR)"= ("BC"/"QR")^2` .........eq5

From eq1 and eq5

`"ar(∆ABC)"/"ar(∆PQR)"= ("AB"/"PQ")^2 = ("BC"/"QR")^2 = ("AC"/"PR")^2`

Hence proved.