Triangles | Exercise 6.2

Question 3

In Fig. 6.18, if LM∥CB\text{LM} \parallel \text{CB} and LN∥CD\text{LN} \parallel \text{CD}, prove that

AMAB=ANAD\dfrac{\text{AM}}{\text{AB}} = \dfrac{\text{AN}}{\text{AD}}

Question diagram 1
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Solution
Understand the Question
  • We are given a figure with quadrilateral ABCD\text{ABCD} divided by diagonal AC\text{AC} into two triangles: ΔABC\Delta \text{ABC} and ΔADC\Delta \text{ADC}.
  • In ΔABC\Delta \text{ABC}, line LM∥CB\text{LM} \parallel \text{CB}, and in ΔADC\Delta \text{ADC}, line LN∥CD\text{LN} \parallel \text{CD}.
  • According to the Basic Proportionality Theorem (BPT), a line drawn parallel to one side of a triangle divides the other two sides in the same ratio.
  • We apply BPT to both triangles with respect to the common side AC\text{AC} and equate the resulting ratios.

Step 1 · Apply BPT in ΔABC\Delta \text{ABC}

In ΔABC\Delta \text{ABC}, we are given LM∥CB\text{LM} \parallel \text{CB}.Diagram 1

By Basic Proportionality Theorem (BPT) AMAB=ALAC…(1)\dfrac{\text{AM}}{\text{AB}} = \dfrac{\text{AL}}{\text{AC}} \quad \dots (1)

Step 2 · Apply BPT in ΔADC\Delta \text{ADC}

In ΔADC\Delta \text{ADC}, we are given LN∥CD\text{LN} \parallel \text{CD}.

By Basic Proportionality Theorem (BPT) ANAD=ALAC…(2)\dfrac{\text{AN}}{\text{AD}} = \dfrac{\text{AL}}{\text{AC}} \quad \dots (2)

Step 3 · Compare Equations

From equations (1)(1) and (2)(2), both ratios are equal to ALAC\dfrac{\text{AL}}{\text{AC}} AMAB=ANAD\dfrac{\text{AM}}{\text{AB}} = \dfrac{\text{AN}}{\text{AD}}

Answer

Hence proved,

AMAB=ANAD\dfrac{\text{AM}}{\text{AB}} = \dfrac{\text{AN}}{\text{AD}}

Common Mistakes
  • Ratio Form Confusion: Standard BPT gives AMMB=ALLC\dfrac{\text{AM}}{\text{MB}} = \dfrac{\text{AL}}{\text{LC}}. Adding 11 to both sides or taking reciprocals gives the whole-length form AMAB=ALAC\dfrac{\text{AM}}{\text{AB}} = \dfrac{\text{AL}}{\text{AC}}, which directly matches what needs to be proved.
  • Incorrect Triangle Association: Mixing up the vertices — ensure LM∥CB\text{LM} \parallel \text{CB} is applied strictly to ΔABC\Delta \text{ABC} and LN∥CD\text{LN} \parallel \text{CD} to ΔADC\Delta \text{ADC} with common side AC\text{AC}.

More questions in Exercise 6.2

Q1

In Fig. 6.17, (i) and (ii), DE∥BCDE \parallel BC. Find ECEC in (i) and ADAD in (ii).

Q2

EE and FF are points on the sides PQPQ and PRPR respectively of a ΔPQR\Delta PQR. For each of the following cases, state whether EF∥QREF \parallel QR :

(i) PE=3.9 cmPE = 3.9\text{ cm}, EQ=3 cmEQ = 3\text{ cm}, PF=3.6 cmPF = 3.6\text{ cm}, and FR=2.4 cmFR = 2.4\text{ cm}

(ii) PE=4 cmPE = 4\text{ cm}, QE=4.5 cmQE = 4.5\text{ cm}, PF=8 cmPF = 8\text{ cm}, and RF=9 cmRF = 9\text{ cm}

(iii) PQ=1.28 cmPQ = 1.28\text{ cm}, PR=2.56 cmPR = 2.56\text{ cm}, PE=0.18 cmPE = 0.18\text{ cm}, and PF=0.36 cmPF = 0.36\text{ cm}

Q3

In Fig. 6.18, if LM∥CB\text{LM} \parallel \text{CB} and LN∥CD\text{LN} \parallel \text{CD}, prove that

AMAB=ANAD\dfrac{\text{AM}}{\text{AB}} = \dfrac{\text{AN}}{\text{AD}}

Q4

In Fig. 6.19, DE∥ACDE \parallel AC and DF∥AEDF \parallel AE. Prove that

BFFE=BEEC\dfrac{\text{BF}}{\text{FE}} = \dfrac{\text{BE}}{\text{EC}}

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