Question

In the figure alongside,

AB = AC

$\angle$ A = ${48}^0$ and

$\angle$ ACD = ${18}^0$

Show that : BC = CD.

Answer 1

Here
∠ A = 48°
and ∠ ACD = 18°
and AB = AC

So,

From base angle theorem

∠ ABC = ∠ ACB

And from angle sum property of triangle we get in triangle ABC

∠ A + ∠ ABC + ∠ ACB = 180° , So

48° + ∠ ABC + ∠ ABC = 180° ( Given ∠ A = 48° and we know ∠ ABC = ∠ ACB )

2 ∠ ABC = 132°

∠ ABC = 66° , So

∠ ABC = ∠ ACB = 66° ------ ( 1 )


And

From angle sum property of triangle we get in triangle ACD

∠ A + ∠ ACD + ∠ ADC = 180° , So

48° +18° + ∠ ADC = 180° ( Given ∠ A = 48° and ∠ ACD = 18° )

∠ ADC = 114°

And

∠ ADC and ∠ BDC are supplementary ,Because these angles are linear pair angles . So

∠ ADC + ∠BDC = 180°

114° + ∠BDC = 180°

∠BDC = 66° ------ ( 2 )

From equation 1 and 2 we get

∠ABC = ∠BDC = 66° , So

∠DBC = ∠BDC , from base angle theorem we get in triangle BCD


BC =CD ( Hence proved )