Question

In the figure alongside,

AB = AC
∠A = 48°  and
∠ACD = 18° 
Show that BC = CD.

Answer 1

In ΔABC,
∠BAC + ∠ACB + ∠ABC = 180°
48° + ∠ACB + ∠ABC = 180°
But ∠ACB = ∠ABC ........[ AB = AC ]
2∠ABC = 180° - 48°
2∠ABC = 132°
∠ABC = 66° = ∠ACB ..........(i)

∠ACB = 66°
∠ACD + ∠DCB = 66°
18° + ∠DCB = 66°
∠DCB = 48° ..........(ii)

Now, In ΔDCB,
∠DBC = 66° .......[ From (i), Since ∠ABC = ∠DBC ]
∠DCB = 48° .......[From (ii)]
∠BDC = 180°- 48° - 66°
∠BDC = 66°
Since ∠BDC = ∠DBC
Therefore, BC = CD
Equal angles have equal sides opposite to them.