Q7 of 13 Page 96

In trapezium ABCD, AD||BC and ∠ABC=∠BCD, let’s prove that ABCD is an isosceles trapezium.


Given a trapezium ABCD in which AD||BC and ∠ABC=∠BCD


Draw AEꞱBC and DFꞱBC


∠ABC=∠BCD (given)…………..(1)


∠AEB=∠DFC=90°………………(2)


Now, in ΔAEB,


∠BAE+∠AEB+∠ABE=180° (sum of all the angles of a Δ)……….(3)


In ΔDFC,


∠FDC+∠DFC+∠DCF=180° (sum of all the angles of a Δ)……….(4)


Subtracting equation (4) from (3)


∠BAE+∠AEB+∠ABE-∠FDC-∠DFC-∠DCF=180°-180°


⇒ ∠BAE-∠FDC+∠AEB-∠DFC+∠ABE-∠DCF=0


⇒ ∠BAE-∠FDC=0 (from (1) and (2))


⇒ ∠BAE=∠FDC………………..(5)


Now, in ΔBAE and ΔCDF,


AE=DF (by construction)


∠BAE=∠FDC (from (5))


∠AEB=∠DFC (each 90° )


∴ ΔBAE ≅ ΔCDF (by ASA rule)


So, AB=DC (by cpct)


Hence, the trapezium is isosceles as AB=CD.


More from this chapter

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 6

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