If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. This is also called SAS (Side-Angle-Side) criterion.

**Construction:** Two triangles ABC and DEF are drawn so that one of the angles of one triangle is equal to one of the angles of another triangle. Moreover, two sides included in that angle of one triangle are proportional to two sides included in that angle of another triangle. This means;

∠ A = ∠ D and `(AB)/(DE)=(AC)/(DF)`

**To Prove:** Δ ABC ∼ Δ DEF

Draw PQ in triangle DEF so that, AB = DP and AC = DF

**Proof:**

`ΔABC≅ΔDPQ`

Because corresponding sides of these two triangles are equal

`(AB)/(DE)=(AC)/(DF)` given

∠ A = ∠ D

Hence; `(AB)/(DE)=(BC)/(EF)` from SSS criterion

Hence;

`(AB)/(DE)=(AC)/(DF)=(BC)/(EF)`

Hence; Δ ABC ∼ Δ DEF proved

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

**Construction:** Two triangles ABC and PQR are drawn so that, ΔABC ∼ Δ PQR.

**To Prove:**

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

Draw AD ⊥ BC and PM ⊥ PR

**Proof:**

`text(ar ABC)=1/2xxBCxxAD`

`text(ar PQR)=1/2xxQRxxPM`

Hence; `text(ar ABC)/text(ar PQR)=(BCxxAD)/(QRxxPM)`

Now, in Δ ABD and Δ PQM;

∠ A = ∠ P, ∠ B = ∠ Q and ∠ D = ∠ M (because Δ ABC ∼ Δ PQR)

Hence; Δ ABD ∼ Δ PQM

Hence; `(AD)/(PM)=(AB)/(PQ)`

Since, Δ ABC ∼ Δ PQR

So, `(AB)/(PQ)=(AC)/(PR)=(BC)/(QR)`

Hence; `text(ar ABC)/text(ar PQR)=(AB)/(PQ)xx(AD)/(PM)`

`=(AB)/(PQ)xx(AB)/(PQ)=(AB^2)/(PQ^2)`

Similarly, following can be proven:

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

If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

**Construction:** Triangle ABC is drawn which is right-angled at B. From vertex B, perpendicular BD is drawn on hypotenuse AC.

**To Prove:** Δ ABC ~ Δ ADB ~ Δ BDC

**Proof:** In Δ ABC and Δ ADB;

∠ ABC = ∠ ADB

∠ BAC = ∠ DAB

∠ ACB = ∠ DBA

From AAA criterion; Δ ABC ~ Δ ADB

In Δ ABC and Δ BDC;

∠ ABC = ∠ BDC

∠ BAC = ∠ DBC

∠ ACB = ∠ DBC

From AAA criterion; Δ ABC ~ Δ BDC

Hence; Δ ABC ~ Δ ADB ~ Δ BDC proved.

**Pythagoras Theorem:** In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

**Construction:** Triangle ABC is drawn which is right angled at B. From vertex B, perpendicular BD is drawn on hypotenuse AC.

**To Prove:** `AC^2 = AB^2 + BC^2`

**Proof:** In Δ ABC and Δ ADB;

`(AB)/(AC)=(AD)/(AB)`

Or, `ACxxAD=AB^2`

Because these are similar triangles (as per previous theorem)

In Δ ABC and Δ BDC;

`(BC)/(AC)=(CD)/(BC)`

Or, `ACxxCD=BC^2`

Adding equations (1) and (2), we get;

`AC xx AD + AC xx CD = AB^2 + BC^2`

Or, `AC(AD + CD) = AB^2 + BC^2`

Or, `AC xx AC = AB^2 + BC^2`

Or, `AC^2 = AB^2 + BC^2`

Proved

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