PSLE In the figure, EFGH is a parallelogram. CHE and DHG are straight lines and CD = DE. ∠CDH = 25° and ∠FGH = 55°.
- Find ∠GHE.
- Find ∠HDE.
(a)
∠GHE
= 180° - 55°
= 125° (Interior angles)
(b)
∠CHD
= ∠GHE
= 125° (Vertically opposite angles)
∠DCH
= 180° - 125° - 25°
= 30° (Angles sum of triangle)
∠DEH = ∠DCH (Isosceles triangle)
∠CDE
= 180° - 30° - 30°
= 120° (Angles sum of triangle)
∠HDE
= 120° - 25°
= 95°
Answer(s): (a) 125°; (b) 95°