Q20 of 153 Page 166

In ΔABC, ∠B = 90° and BD ⊥ AC. Prove that ∠ABD = ∠ACB.

Let ∠ABD = x and ∠ACB = y

According to question,

∠B = 90^O

In triangle BDC, we have,

∠BDC = 90^O

∠DBC = (90 – x)^O

∠BDC + ∠DBC + ∠DCB = 180^O

90^O + (90 – x)^O + y = 180^O

180^O – x + y = 180^O

x = y

So,

∠ABD = ∠ACB

In the given figure AB || CD, ∠APQ = 50° and ∠PRD = 120°. Find ∠QPR.

In the given figure, BE is the bisector of ∠B and CE is the bisector of ∠ACD.

Prove that

In ΔABC, sides AB and AC are produced to D and E respectively. BO and CO are the bisectors of ∠CBD and ∠BCE respectively. Then, prove that

Of the three angles of a triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.