Question

As shown in the diagram. BE
and CF bisect ZABD and
ZACD respectively. BE and
CF intersect at G. Given that
ZBDC = 150° and ZBGC = 100°,
find angle ZA
А
F
E
G
D
B
С​

Answer 1

Answer:

ANSWER

⇒∠CBE+∠BCF=180° ( Angle between two parallel lines)

2

∠CBE

+

2

∠BCF

=90° (From above)

∠CBG=

2

∠CBE

and ∠BCG=

2

∠BCF

So, ∠CBG+∠BCG=90°

Now, ∠BGC+∠CBG+∠BCG=180°

∠BGC+90°=180°

∠BGC=90°