Question

(a) State and Prove "Basic Proportionality Theorem".

(b) In the given figure, CM and RN are respectively, the medians of $△ A BC$ and $△ PQR$ . If $△ A BC \sim △ PQR$ , prove that:

(i) $△ A MC \sim △ PNR$

(ii) $∠ BCM = ∠ QRN$

(iii) $△ BMC \sim △ QNR$

Approach

Basic Proportionality Theorem: If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio. Given $△ A BC$ with DE $∥$ BC, construct perpendiculars from D and E to the other sides. Using area ratios, show $\frac{A D}{D B} = \frac{A E}{EC}$ .

Step-by-step working

1
Statement: If DE $∥$ BC in $△ A BC$ , then $\frac{A D}{D B} = \frac{A E}{EC}$
1 mark
2
Given: $△ A BC$ , DE $∥$ BC. To Prove: $\frac{A D}{D B} = \frac{A E}{EC}$ . Construction: Draw DM $⊥$ AC, EN $⊥$ AB, join BE and CD
1 mark
3
$\frac{ar ( △ A D E )}{ar ( △ D BE )} = \frac{\frac{1}{2} \times A D \times EN}{\frac{1}{2} \times D B \times EN} = \frac{A D}{D B}$
1 mark
4
$\frac{ar ( △ A D E )}{ar ( △ EC D )} = \frac{\frac{1}{2} \times A E \times D M}{\frac{1}{2} \times EC \times D M} = \frac{A E}{EC}$
1 mark
5
Since DE $∥$ BC, $ar (△ D BE) = ar (△ EC D)$ , so $\frac{A D}{D B} = \frac{A E}{EC}$
1 mark

Concepts used

basic proportionality theoremprovesimilar trianglesmedianstheorem

Approach

Step-by-step working

Concepts used

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