What conclusion can be drawn from each part of Fig. 2.41, if

(a) DB is the bisector of ∠ADC?

(b) BD bisects ∠ABC?

(c) DC is the bisector of ∠ADB, CA ⊥ DA and CB ⊥ DB?

a) Since, DB is the bisector of ∠ ADC.

This means that DB divides ∠ ADC into 2 equal parts.

∴ ∠ ADB = ∠ BDC

b) Since, BD bisects ∠ ABC.

This means that, ∠ ABD = ∠ DBC.

c) Since, DC is the bisector of ∠ ADB

∴ ∠ ADC = ∠ CDB ………….. (1)

Also, it is given that,

∠ CAD = ∠ CBD = 90° ……….. (2)

Also, we know that sum of interior angles of a triangle is equal to 180°.

∴ In Δ ACD,

∠ ACD+∠ CDA+ ∠ DAC =180° …. (3)

In Δ BCD,

∠ BCD+∠ CDB+∠ DBC =180° …. (4)

From (1), (2), (3), (4) we get,

∠ ACD = ∠ BCD