Let ΔABC be a triangle such that ∠B = 70° and∠C = 40° . suppose

1.	Let ΔABC be a triangle such that ∠B = 70° and∠C = 40° . suppose D is a point on BC such that AB = AD. Prove that AB > CD.
Answer» In Δ ABD, AB = AD ∴∠ABD = ∠ APB = 70° ∠APB +∠ ADC = 180° [Linear pair] 70° +∠ADC = 180° ∠ADC = 180° – 70° ∠ADC = 110° In Δ ADC, ∠DAC = 180° – ( 110° + 40°) = 180°-150 ° ∠DAC = 30° In Δ ADC, ∠ACD ∠DAC AD > CD but AB = AD ∴ AB > CD

Let ΔABC be a triangle such that ∠B = 70° and∠C = 40° . suppose D is a point on BC such that AB = AD. Prove that AB > CD.

Answer»

In Δ ABD, AB = AD

∴∠ABD = ∠ APB = 70°

∠APB +∠ ADC = 180° [Linear pair]

70° +∠ADC = 180°

∠ADC = 180° – 70°

∠ADC = 110°

In Δ ADC,

∠DAC = 180° – ( 110° + 40°)

= 180°-150 °

∠DAC = 30°

In Δ ADC,

∠ACD ∠DAC

AD > CD

but AB = AD

∴ AB > CD