10 Math Triangle

Finish Line & Beyond TRIANGLES Similar Figures All congruent figures are similar but the similar figures need not be congruent. Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). This again emphasises that two polygons of the same number of sides are similar, if (i) all the corresponding angles are equal and (ii) all the corresponding sides are in the same ratio (or proportion). If one polygon is similar to another polygon and this second polygon is similar to a third polygon, then the first polygon is similar to the third polygon. Similarity of Triangles Two triangles are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Proof : We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.

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N

D

E

C

B

We need to prove that,

AD AE = DB EC

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Finish Line & Beyond ⊥ AC and EN ⊥ AB. 1 1 × AD × EN Now, area of Δ ADE = × base × height = 2 2 1 Similarly ar(BDE) = × BD × EN 2 1 ar(ADE) = × AE × DM 2 1 ar(DCE) = × EC × DM 2 ar ( ADE ) AD × EN AD So, = = ---------------------------------- (1) ar ( BDE ) BD × EN BD ar ( ADE ) AE × DM AE = = ---------------------------------- (2) ar ( DCE ) EC × DM EC Now, as ∆ BDE and ∆ DCE are on the same side of DE Let us join BE and CD and then draw DM

So, ar(BDE) = ar(DCE)

---------------------------------- (3)

So, from three above equations it is proved that

AD AE = BD EC Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. Proof: Draw a triangle ABC and draw a line DE such as it divides AB and AC in the same ratio.

A

D

B

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E

C

Finish Line & Beyond In

∆ ABC it is given that

AD AE = AB EC

Prove DE ll BC Proof: ∠ AED + ∠ DEC = 180° In Rectangle DECB, ∠ DEC + ∠ ECB = 180° So, ∠ AED = ∠ ECB As per theorem of parallel lines angles on the same side of a tansverse section line of two parallel lines are always equal. So, it is proved that DE ll BC AAA (Angle-Angle-Angle) Theorem: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

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In ∆ ABC and ∆ DEF, ∠ A= ∠ D ∠ B= ∠ E ∠ C= ∠ F Let us draw a line PQ so that DP = AB and DQ = AC So, ∆ ABC ≅ ∆ DPQ So, ∠ B = ∠ P = ∠ E and PQ ll EF, because by making a similar triangle like ABC the line PQ bisects lines DE and DF in the same ratio.

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Finish Line & Beyond AB BC = DE DF So, ∆ ABC ≈ ∆ DEF proved Hence,

Note: If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. SSS (Side-Side-Side)Theorem: If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar. Proof: This theorem can be proved by using same method as in previous theorem. SAS(Side-Angle-Side)Theorem: 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. Areas of Similar Triangles Theorem: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Proof : We are given two triangles ABC and PQR such that Δ ABC ~ Δ PQR

D

A

B

M

C E

ar(ABC) =

1 1 × BC × AM and ar(DEF) = × EF × DN 2 2

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N

F

Finish Line & Beyond ar ( ABC ) BC × AM = --------------------------------------------- (1) ar ( DEF ) EF × DN ar ( ABC )  AB²   BC²   AC²  We need to Prove =   =  =  ar ( DEF )  DE²   EF²   DF²  So,

BC AB AC = = EF DE DF As AM ⊥ BC and DN ⊥ EF so ∆ABM ≈ ∆DEN BC AB AC AM So, = = = ----------------------------------------------------------- (2) EF DE DF DN As triangles are similar so

Putting this into equation (1) we get,

ar ( ABC ) BC × BC = proved. ar ( DEF ) EF × EF Theorem: 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.

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∆ABC is a right angle triangle and BD ⊥ AC Proof: As per SAS theorem, In ∆ADB and ∆DCB, and ∆ ABC ∠ ADB = ∠ BDC = ∠ ABC ∠ ADB + ∠ DBA + ∠ BAD = 180°

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∠ ∠ ∠ ∠

BDC + ∠ BCD + ∠ CBD = 180° ABC + ∠ ACB + ∠ BAC = 180° BAC = ∠ BAD ACB = ∠ BCD So, ∠ BAC = ∠ DBC And ∠ BCD = ∠ ABD So, ∆ADB

Finish Line & Beyond

≈ ∆DCB ≈ ∆ ABC proved

Theorem: In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. This is converse of Pythagoras Theorem

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