Question

Let ABC be an actual angled triangle ; AD be bisector of ∠BAC with D on BC and BE be the altitude from B on AC. Show that ∠CED > 45⁰ 

Answer 1

Explanation:

Draw DL perpendicular to AB; DK perpendicular to AC and DM perpendicular to BE. Then EM = DK. Since AD bisects ∠A, we observe that ∠BAD = ∠KAD. 

Thus in triangles ALD and AKD, we see that ∠LAD = ∠KAD; ∠AKD = 90° = ∠ALD; and AD is common. 

Hence triangles ALD and AKD are congruent, giving DL = DK. 

But DL >DM, since BE lies inside the triangle (by acuteness property). 

Thus EM > DM. This implies that ∠EDM >∠DEM = 90° – ∠EDM. We conclude that ∠EDM > 45°. Since ∠CED = ∠EDM, the result follows.

Alternate Solution: