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Ex 7.3, 1 - Triangle ABC and DBC are two isosceles triangles - Ex 7.3

  1. Chapter 7 Class 9 Triangles
  2. Serial order wise
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Ex 7.3,1 ∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure ). If AD is extended to interest BC at P, show that ∆ ABD ≅ ∆ ACD Given: ∆ABC is isosceles, AB = AC Also, ∆DBC is isosceles, DB = DC To prove: ∆ ABD ≅ ∆ ACD Proof : In ∆ABD and ∆ DBC, we have AB = AC BD = DC AD = AD ∆ ABD ≅ ∆ ACD Ex 7.3,1 ∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure ). If AD is extended to interest BC at P, show that (ii) ∆ ABP ≅ ∆ ACP From part (i), ∆ ABD ≅ ∆ ACD So, ∠BAP = ∠PAC In ∆ABP and ∆ ACP, AB = AC ∠BAP = ∠PAC AP = AP ∆ ABP ≅ ∆ ACP. Ex 7.3,1 ∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure ). If AD is extended to interest BC at P, show that (iii)AP bisects ∠𝐴 as well as ∠𝐷. To prove: ∠𝐵𝐴𝐷 = ∠𝐶𝐴𝐷 & ∠𝐵𝐷𝑃 = ∠𝐶𝐷𝑃 Proof: From part (i) , ∆ ABD ≅ ∆ ACD So, ∠𝐵𝐴𝐷 = ∠𝐶𝐴𝐷. Hence AP bisects ∠ A For ∠𝐵𝐷𝑃 = ∠𝐶𝐷𝑃, we will first prove Δ BDP ≅ Δ CDP From part(ii) , ∆ ABP ≅ ∆ ACP BP = CP In ∆ BDP and CDP, we have BD = CD BP = CP DP = DP So, ∆ BDP ≅ ∆ CDP ⇒ ∠𝐵𝐷𝑃 = ∠𝑃𝐷𝐶 Thus, AP bisects ∠ D . Ex 7.3,1 ∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure ). If AD is extended to interest BC at P, show that (iv) AP is the perpendicular bisector of BC To prove: BP = CP & ∠𝐴𝑃𝐵 = ∠𝐴𝑃𝐶 = 90∘ Proof From part(ii) , ∆ ABP ≅ ∆ ACP BP = CP ∠𝐴𝑃𝐵 = ∠𝐴𝑃𝐶 Since BC is a line, ∴ ∠𝐴𝑃𝐵 + ∠𝐴𝑃𝐶 = 180∘ ∠𝐴𝑃𝐵 + ∠𝐴𝑃𝐵 = 180∘ 2∠𝐴𝑃𝐵 = 180∘ ∠𝐴𝑃𝐵 = (180∘)/2 ∠𝐴𝑃𝐵 = 90∘ So, ∠𝐴𝑃𝐵 = ∠𝐴𝑃𝐶 = 90∘ Since, BP = CP & ∠𝐴𝑃𝐵 = ∠𝐴𝑃𝐶 = 90∘ ⇒ AP is perpendicular bisector of BC.

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