PSLE In the figure, EFGH is a parallelogram. CHE and DHG are straight lines and CD = DE. ∠CDH = 23° and ∠FGH = 56°.
- Find ∠GHE.
- Find ∠HDE.
(a)
∠GHE
= 180° - 56°
= 124° (Interior angles)
(b)
∠CHD
= ∠GHE
= 124° (Vertically opposite angles)
∠DCH
= 180° - 124° - 23°
= 33° (Angles sum of triangle)
∠DEH = ∠DCH (Isosceles triangle)
∠CDE
= 180° - 33° - 33°
= 114° (Angles sum of triangle)
∠HDE
= 114° - 23°
= 91°
Answer(s): (a) 124°; (b) 91°