Ex 5.4, 3

Transcript

Ex 5.4, 3 Solve the previous question using the Baudhāyana-Pythagoras theorem. Given: : 𝐶𝐸⊥𝐴𝐵 and 𝐶𝐻⊥𝐺𝐹. And, distances are equal i.e. CE = CH To Prove: AB = GF Proof: First we prove that AE is half of AB, and GH is half of GF Then, we apply Baudhāyana-Pythagoras in ∆ CAE & in ∆ CGH We know that Perpendicular from the center to the chord, bisects the chord For Chord AB Since CE ⊥ AB, ∴ E bisects AB. So, we can write AE = BE = 1/2AB For Chord GF Since CH ⊥ GF ∴ H bisects GF. So, we can write GH = FH = 1/2GF Since ∆ AEC and ∆GHC are right angled triangles Applying Baudhāyana–Pythagoras theorem to each In ∆ AEC By Pythagoras theorem 〖𝑨𝑪〗^𝟐=〖𝑪𝑬〗^𝟐+〖𝑨𝑬〗^𝟐 Putting AC = Radius = r 𝒓^𝟐=〖𝑪𝑬〗^𝟐+〖𝑨𝑬〗^𝟐 In ∆ GHC By Pythagoras theorem 〖𝑪𝑮〗^𝟐=〖𝑪𝑯〗^𝟐+〖𝑮𝑯〗^𝟐 Putting OA = Radius = r 𝒓^𝟐=〖𝑪𝑯〗^𝟐+〖𝑮𝑯〗^𝟐 Comparing (3) & (4) 〖𝑪𝑬〗^𝟐+〖𝑨𝑬〗^𝟐=〖𝑪𝑯〗^𝟐+〖𝑮𝑯〗^𝟐 Given that CE = CH 〖𝑪𝑯〗^𝟐+〖𝑨𝑬〗^𝟐=〖𝑪𝑯〗^𝟐+〖𝑮𝑯〗^𝟐 𝐴𝐸^2 = 〖𝐶𝐻〗^2+〖𝐺𝐻〗^2−𝐶𝐻^2 〖𝑨𝑬〗^𝟐 = 〖𝑮𝑯〗^𝟐 Cancelling squares AE = GH Putting AE = 𝟏/𝟐AB & GH = 𝟏/𝟐GF 1/2 AB = 1/2 GF Cancelling 1/2 both sides AB = GF Hence proved