Given : AB || DE and BC || EF. Prove that : (i) AD/DG = CF/FG (ii) △DFG ~ △ACG.

(i) In △ABG, DE || AB. So, by basic proportionality theorem we get, .....(1) In △BCG, EF || BC. So, by basic proportionality theorem we get, .....(2) From (1) and (2) we get, Hence, proved that (ii) In △DFG and △ACG, ⇒ [Proved above] ⇒ ∠DGF = ∠AGC [Common angle] ∴ △DFG ~ △ACG [By SAS] Hence, proved that △DFG ~ △ACG.

Mathematics

Given : AB || DE and BC || EF. Prove that :
(i) $\dfrac{AD}{DG} = \dfrac{CF}{FG}$
(ii) △DFG ~ △ACG.

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Answer

(i) In △ABG, DE || AB.

So, by basic proportionality theorem we get,

$\dfrac{DG}{AD} = \dfrac{GE}{BE}$ …..(1)

In △BCG, EF || BC.

So, by basic proportionality theorem we get,

$\dfrac{GE}{BE} = \dfrac{FG}{CF}$ …..(2)

From (1) and (2) we get,

$\Rightarrow \dfrac{DG}{AD} = \dfrac{FG}{CF} \\[1em] \Rightarrow \dfrac{AD}{DG} = \dfrac{CF}{FG}.$

Hence, proved that $\dfrac{AD}{DG} = \dfrac{CF}{FG}.$

(ii) In △DFG and △ACG,

⇒ $\dfrac{AD}{DG} = \dfrac{CF}{FG}$ [Proved above]

⇒ ∠DGF = ∠AGC [Common angle]

∴ △DFG ~ △ACG [By SAS]

Hence, proved that △DFG ~ △ACG.

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Mathematics

Given : AB || DE and BC || EF. Prove that :
(i) $\dfrac{AD}{DG} = \dfrac{CF}{FG}$
(ii) △DFG ~ △ACG.

Similarity

Answer

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Given : AB || DE and BC || EF. Prove that :
(i) $\dfrac{AD}{DG} = \dfrac{CF}{FG}$
(ii) △DFG ~ △ACG.

In the given figure, ABC is a right angled triangle with ∠BAC = 90°.
(i) Prove that : △ADB ~ △CDA.
(ii) If BD = 18 cm and CD = 8 cm, find AD.
(iii) Find the ratio of the area of △ADB is to area of △CDA.

ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :
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(iii) Find, area of △ADE : area of quadrilateral BCED.

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(i) length of AB
(ii) area of △ABC