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Triangles - Prove Basic Proportionality Theorem and apply it in geometric problems

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Basic Proportionality Theorem (Thales' Theorem) states that 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. In ΔABC\Delta ABC, if DE∥BCDE \parallel BC, then ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}.

Triangle ABC with line DE parallel to BC intersecting AB at D and AC at E.
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The Converse of BPT states that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. If ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}, then DE∥BCDE \parallel BC.

A triangle showing segments on sides in equal ratios implying parallel lines.
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Corollaries of BPT: The ratios can also be expressed comparing parts to the whole side: ADAB=AEAC\frac{AD}{AB} = \frac{AE}{AC} and DBAB=ECAC\frac{DB}{AB} = \frac{EC}{AC}. This is useful when the full length of a side is known.

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Internal Bisector Theorem: A special application related to proportionality where the internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle.

📐Formulae

Basic Proportionality Theorem: ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}

Converse of BPT: If ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}, then DE∥BCDE \parallel BC

Corollary 1 (Whole side to upper segment): ABAD=ACAE\frac{AB}{AD} = \frac{AC}{AE}

Corollary 2 (Whole side to lower segment): ABDB=ACEC\frac{AB}{DB} = \frac{AC}{EC}

Extended Ratio (Similarity): ADAB=AEAC=DEBC\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}

💡Examples

Problem 1:

In ΔABC\Delta ABC, DE∥BCDE \parallel BC. If AD=1.5AD = 1.5 cm, DB=3DB = 3 cm, and AE=1AE = 1 cm, find the length of ECEC.

Solution:

Given DE∥BCDE \parallel BC, by Basic Proportionality Theorem: ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC} Substitute the given values: 1.53=1EC\frac{1.5}{3} = \frac{1}{EC} Simplify the fraction on the left: 12=1EC\frac{1}{2} = \frac{1}{EC} Cross-multiplying gives: EC=2 cmEC = 2 \text{ cm}

Explanation:

We use the direct ratio provided by Thales' Theorem because the line DEDE is parallel to the base BCBC and we need to find a segment on one of the divided sides.

Problem 2:

In ΔABC\Delta ABC, DE∥BCDE \parallel BC. If AD=xAD = x, DB=x−2DB = x - 2, AE=x+2AE = x + 2, and EC=x−1EC = x - 1, find the value of xx.

Solution:

Using BPT: ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC} Substitute the algebraic expressions: xx−2=x+2x−1\frac{x}{x - 2} = \frac{x + 2}{x - 1} Cross-multiply to solve for xx: x(x−1)=(x+2)(x−2)x(x - 1) = (x + 2)(x - 2) x2−x=x2−4x^2 - x = x^2 - 4 Subtract x2x^2 from both sides: −x=−4-x = -4 x=4x = 4

Explanation:

This problem applies BPT to solve an algebraic equation. Note that (x+2)(x−2)(x+2)(x-2) simplifies to x2−4x^2 - 4 using the identity (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2.

Problem 3:

In the given figure, ST∥QRST \parallel QR. If PS=4PS = 4 cm, PQ=10PQ = 10 cm, and PR=15PR = 15 cm, find the length of PTPT.

Triangle PQR with line ST parallel to QR. PS=4 and PQ=10.

Solution:

  1. Identify the given values: PS=4PS = 4 cm, PQ=10PQ = 10 cm.
  2. Calculate SQ=PQ−PS=10−4=6SQ = PQ - PS = 10 - 4 = 6 cm.
  3. By Basic Proportionality Theorem, since ST∥QRST \parallel QR, PSSQ=PTTR\frac{PS}{SQ} = \frac{PT}{TR}.
  4. Alternatively, use the corollary: PSPQ=PTPR\frac{PS}{PQ} = \frac{PT}{PR}.
  5. Substituting the values: 410=PT15\frac{4}{10} = \frac{PT}{15}.
  6. 10×PT=4×15  ⟹  10×PT=6010 \times PT = 4 \times 15 \implies 10 \times PT = 60.
  7. PT=6010=6PT = \frac{60}{10} = 6 cm.

Explanation:

We use the corollary of the Basic Proportionality Theorem which relates the upper segment of the side to the total length of the side.

Problem 4:

In ΔABC\Delta ABC, DD and EE are points on sides ABAB and ACAC respectively such that AD=8AD = 8 cm, DB=12DB = 12 cm, AE=6AE = 6 cm and EC=9EC = 9 cm. Prove that DE∥BCDE \parallel BC.

Triangle ABC with segments AD=8, DB=12, AE=6, EC=9.

Solution:

  1. Calculate the ratio of segments on side ABAB: ADDB=812=23\frac{AD}{DB} = \frac{8}{12} = \frac{2}{3}.
  2. Calculate the ratio of segments on side ACAC: AEEC=69=23\frac{AE}{EC} = \frac{6}{9} = \frac{2}{3}.
  3. Compare the ratios: Since ADDB=AEEC=23\frac{AD}{DB} = \frac{AE}{EC} = \frac{2}{3}, the line DEDE divides the two sides ABAB and ACAC in the same ratio.
  4. By the Converse of Basic Proportionality Theorem, DEDE must be parallel to BCBC.

Explanation:

This problem applies the Converse of BPT. Since the ratios of the split segments are equal, the line creating the split is parallel to the base.