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

Q33(a): Basic Proportionality Theorem (Thales): If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally. Given: In $△ A BC$ , $D E ∥ BC$ . To prove: $\frac{A D}{D B} = \frac{A E}{EC}$ . Construct perpendiculars from $D$ and $E$ to $A C$ and $A B$ respectively. Use area ratios of triangles sharing the same base and height to establish the equality. | (b)(i): Since $△ A BC \sim △ PQR$ , corresponding sides are proportional: $\frac{A B}{PQ} = \frac{A C}{PR}$ . Medians divide sides in the same ratio, so $\frac{A M}{PN} = \frac{A B}{PQ}$ . Also $∠ A = ∠ P$ . By SAS similarity, $△ A MC \sim △ PNR$ . | (b)(ii): From $△ A MC \sim △ PNR$ (part i), corresponding angles are equal: $∠ A CM = ∠ PRN$ . Also, since $△ A BC \sim △ PQR$ , $∠ A CB = ∠ PRQ$ . Subtracting: $∠ A CB - ∠ A CM = ∠ PRQ - ∠ PRN$ , so $∠ BCM = ∠ QRN$ . | (b)(iii): Since $△ A BC \sim △ PQR$ , $\frac{A B}{PQ} = \frac{BC}{QR}$ . Also $\frac{BC}{QR} = \frac{\frac{1}{2} A B}{\frac{1}{2} PQ} = \frac{BM}{QN}$ . Also $∠ B = ∠ Q$ . By SAS similarity, $△ BMC \sim △ QNR$ .

Step-by-step working

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Q33(a) Statement: 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.
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Q33(a) Given: In $△ A BC$ , $D E ∥ BC$ . To prove: $\frac{A D}{D B} = \frac{A E}{EC}$ . Construction: Draw $D M ⊥ A C$ and $EN ⊥ A B$ . Join $BE$ and $C D$ .
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Q33(a) Proof: $\frac{a r ( △ A D E )}{a r ( △ D BE )} = \frac{\frac{1}{2} A D \cdot EN}{\frac{1}{2} D B \cdot EN} = \frac{A D}{D B}$ .
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Q33(a) Similarly, $\frac{a r ( △ A D E )}{a r ( △ EC D )} = \frac{\frac{1}{2} A E \cdot D M}{\frac{1}{2} EC \cdot D M} = \frac{A E}{EC}$ .
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Q33(a) Since $△ D BE$ and $△ EC D$ lie on the same base $D E$ and between the same parallels $BC$ and $D E$ , $a r (△ D BE) = a r (△ EC D)$ . Hence $\frac{A D}{D B} = \frac{A E}{EC}$ .
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(b)(i) Since $△ A BC \sim △ PQR$ , $\frac{A B}{PQ} = \frac{A C}{PR}$ .
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(b)(i) Medians: $\frac{A C}{PR} = \frac{\frac{1}{2} A B}{\frac{1}{2} PQ} = \frac{A M}{PN}$ , and $∠ A = ∠ P$ .
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(b)(i) By SAS similarity criterion, $△ A MC \sim △ PNR$ .
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(b)(ii) From part (i), $△ A MC \sim △ PNR$ , so $∠ A CM = ∠ PRN$ .
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(b)(ii) Also $∠ A CB = ∠ PRQ$ (from $△ A BC \sim △ PQR$ ). Subtracting: $∠ BCM = ∠ QRN$ .
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(b)(iii) Since $△ A BC \sim △ PQR$ , $\frac{A B}{PQ} = \frac{BC}{QR}$ .
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(b)(iii) Medians: $\frac{BC}{QR} = \frac{\frac{1}{2} A B}{\frac{1}{2} PQ} = \frac{BM}{QN}$ , and $∠ B = ∠ Q$ .
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(b)(iii) By SAS similarity criterion, $△ BMC \sim △ QNR$ .
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Concepts used

basic proportionality theoremthales theoremprovesimilar trianglesmediansprove similarity

Approach

Step-by-step working

Concepts used

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