Ex 7.1, 7 - AB is a line segment and P is its mid-point - ASA/AAS

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Ex7.1, 7 AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that ΔDAP ≅ ΔEBP (ii) AD = BE Given: P is the mid point of AB, So, AP = BP ∠BAD = ∠ABE ∠EPA = ∠DPB To prove: (i) ΔDAP ≅ ΔEBP (ii) AD = BE Proof: Since ∠EPA = ∠DPB We add ∠DPE both sides ∠EPA + ∠DPE = ∠DPB + ∠DPE ∠DPA = ∠EPB In  ∆DAP and  ∆EBP, ∠DAP = ∠EBP AP = BP ∠DPA = ∠EPB ∴ ΔDAP ≅ ΔEBP ∴ AD = BE

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.