## Areas of Two Similar Triangles:

**Theorem 1:** Prove that the ratio of the areas of two similar triangles is equal to the ratio of the square of any two corresponding sides.

Given: △ABC ~ △DEFSo, AB/DE = BC/EF = AC/DF Also, ∠A = ∠D, ∠B = ∠E, ∠C = ∠F To prove: area △ABC/area △DEF = AB^{2}/DE^{2 }= BC^{2}/EF^{2} = AC^{2}/DF^{2}Construction: Through A draw AP ⊥ BC and through D draw DQ ⊥ EF.Proof: area (△ABC) = (1/2) BC x AP; area (△DEF) = (1/2) EF x DQ.Thus, area △ABC/area △DEF = [(1/2) BC x AP]/[(1/2) EF x DQ] ⇒ area △ABC/area △DEF = (BC/EF) x (AP/DQ) ……….(i) In △APB and △DQE, ∠1 = ∠2 = 90° [By construction] ∠3 = ∠4 [given] ∴ △APB ~ △DQE [AA corollary] ∴ AB/DE = AP/DQ = BP/EQ ⇒ AB/DE = AP/DQ ……….(ii) Also, AB/DE = BC/EF [given] ……….(iii) From (ii) and (iii), we get AP/DQ = BC/EF ……….(iv) Putting the value of AP/DQ from (iv) in (i), we get- area △ABC/area △DEF = (BC/EF) x (BC/EF) = BC ^{2}/EF^{2} ……….(v)Similarly. it can also be proved that area △ABC/area △DEF = AB ^{2}/DE^{2} ……….(vi)and area △ABC/area △DEF = AC ^{2}/DF^{2} ……….(vii)From (v), (vi) and (vii), we obtain area △ABC/area △DEF = AB ^{2}/DE^{2} = BC^{2}/EF^{2} = AC^{2}/DF^{2} |

## Some More Theorems On Areas:

**Theorem 1:** The areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.

Given: △ABC ~ △DEF and AL ⊥ BC, DM ⊥ EFTo prove: area (△ABC)/area (△DEF) = AL^{2}/DM^{2}Proof: Since, the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.∴ area (△ABC)/area (△DEF) = AB ^{2}/DE^{2} ……….(i)In △ALB and △DME, ∠ALB = ∠DME = 90° and ∠B = ∠E [ ∵ △ABC ~ △DEF]∴ △ALB ~ △DME [By AAcorollary] So, AB/DE = AL/DM and therefore, AB ^{2}/DE^{2} = AL^{2}/DM^{2} ……….(ii)Substituting for AB ^{2}/DE^{2} in (i), we getarea (△ABC)/area (△DEF) = AL ^{2}/DM^{2} |

**Theorem 2:** The areas of two similar triangles are in the ratio of the squares of the corresponding medians.

Given: △ABC ~ △DEF and AP, DQ are their medians.To prove: area (△ABC)/area (△DEF) = AP^{2}/DQ^{2}Proof: Since, the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.∴ area (△ABC)/area (△DEF) = AB ^{2}/DE^{2} ……….(i)Now, △ABC ~ △DEF ⇒ AB/DE = BC/EF = 2BP/2EQ = BP/EQ ……….(ii) Thus, AB/DE = BP/EQ and ∠B = ∠E [∵ △ABC ~ △DEF] ∴ △APB ~ △DQE [By SAS similarity] So, BP/EQ = AP/DQ ……….(iii) From (ii) and (iii), we get AB/DE = AP/DQ and therefore, AB ^{2}/DE^{2} = AP^{2}/DQ^{2} ……….(iv)∴ area (△ABC)/area (△DEF) = AP ^{2}/DQ^{2} [from (i) and (iv)] |

**Theorem 3:** The areas of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments.

Given: △ABC ~ △DEF and AX, DY are the bisectors of ∠A and ∠D respectively.To prove: area (△ABC)/area (△DEF) = AX^{2}/DY^{2}Proof: Since, the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides, we havearea (△ABC)/area (△DEF) = AB ^{2}/DE^{2} ……….(i)Now, △ABC ~ △DEF ⇒ ∠A = ∠D ⇒ (1/2) ∠A = (1/2) ∠D ⇒ ∠BAX = ∠EDY ……….(ii) ∴ In △ABX and △DEY, from (ii) ⇒ ∠BAX = ∠EDY and ∠B = ∠E [∵ △ABC ~ △DEF] ∴ △ABX ~ △DEF [By AA corollary] ∴ AB/DE = AX/DY and therefore, AB ^{2}/DE^{2} = AX^{2}/DY^{2}Substituting AX ^{2}/DY^{2} for AB^{2}/DE^{2 }from (iii) in (i), we getarea (△ABC)/area (△DEF) = AX ^{2}/DY^{2} |