Question

The angle bisectors of a trapezium ABCD intersect at
points E, F, G and H. If ZADC = 50°, ZBCD = 70°,
find the angles of EFGH.​

Answer 1

Let AB || DC

Angle ADC = 50° (given)

Therefore, angle BAD = 180°-50° = 130°

Angle BCD = 70° (given)

Angle CBA = 180°-70° = 110°

Therefore angle E = 180°-(130+110/2)° = 60°

Now,

Angle F = 180°-(Angle A+Angle D/2)°

= 180°-(130+50/2)° = 90°

Now,

Angle G = 180° -(Angle C+Angle D)°

= 180° -(70°+50°/2) = 120°

Then,

Angle H = 180°-(Angle B+Angle C/2)°

=180°-(110+70/2)° = 180°-90° = 90°

Ans) Angle E= 60°, Angle F = 90°, Angle G = 120°, Angle H = 90°

Answer 2

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