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महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

In the given figure, seg AC and seg BD intersect each other in point P and APCPBPDPAPCP=BPDP. Prove that, ∆ABP ~ ∆CDP. - Geometry Mathematics 2

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प्रश्न

In the given figure, seg AC and seg BD intersect each other in point P and `"AP"/"CP" = "BP"/"DP"`. Prove that, ∆ABP ~ ∆CDP.

बेरीज

उत्तर

Given: Seg AC and seg BD intersect each other in point P and `"AP"/"CP" = "BP"/"DP"`.

To prove: ∆ABP ~ ∆CDP

Proof: In ∆ABP and ∆CDP,

`"AP"/"CP" = "BP"/"DP"`   ...(Given)

∠APB ≅ ∠CPD       ...(vertically opposite angles)

By SAS test of similarity,

∆ABP ~ ∆CDP 

Hence Proved.

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पाठ 1: Similarity - Practice Set 1.3 [पृष्ठ २२]

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बालभारती Geometry (Mathematics 2) [English] 10 Standard SSC Maharashtra State Board
पाठ 1 Similarity
Practice Set 1.3 | Q 8 | पृष्ठ २२

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