Question

give answers of 3 and 5​

Answer 1

Answer:

Step-by-step explanation:  

Question 3- Given In the figure, OD ⊥OE,OD and OE and bisectors of ∠AOC and ∠BOC.

To show Points A, O and B are collinear i.e., AOB is a straight line.

Proof Since, OD and OD and OE bisect angles ∠AOC and ∠BOC, repectively.

∴ ∠AOC=2∠DOC  

and  ∠COB=2∠COE

On adding Equ. (i) and (ii), we get

and ∠COB=2∠COE

⇒ ∠AOC+∠COB=2∠(DOC+∠COE)

⇒ ∠AOC+∠COB=2∠DOC

⇒ ∠AOC+∠COB=2×90∘

⇒ ∠AOC+∠COB=180∘

∴ ∠AOB=180∘

So, ∠AOCand∠COB are forming linear pair.

Also, AOB is a straight

Hence, point A, O and B are collinear.

Question 5-  Given In Δ ABC, produce BC to DC and the bisectors of ∠ABC and ∠ACD meet at point T.

To prove ∠BTC = 1/2 ∠BAC

Proof In ΔABC, ∠C is an exterior angle.

∴ ∠ACD = ∠ABC+ ∠CAB[exterior angle of a triangle is equal to the sum of two opposite angles]

⇒  1/2 ∠ACD =1/2 ∠CAB + 1/2 ∠ACB       [dividing both side by 2]

⇒  ∠TCD = 1/2 ∠ACB + 1/2∠ABC     .......(1)

[∴ CT isabisec→rof∠ACD⇒ 1/2 ∠ACD = ∠TCD]

In ΔBTC, ∠TCD = ∠BTC + C ∠CBT  [exterior angles of a equal to the sum of two opposite interior angles]

⇒∠TCD =∠BTC + 1/2∠ABC....(ii) [∴ BT bisects of∠ABC⇒ ∠CBT = 12 ∠ABC]

From Eqs. (i) and (ii) ,

1/2 ∠CAB + 1/2 ∠ABC = ∠BTC + 1/2 ∠ABC

⇒ ∠BTC = 1/2 ∠CAB

or ∠BTC = 1/2 ∠ABC