Question

(11)
2HCD
" LDC
(111) ZABU
14. In Figure, AB is a diameter of a circle with centre O.
If ZBOC = 30° and ZCOD = 50°, find
D
C
50°
30°
B.
A
O
(i) ZBAC (ii) ZCAD
(iii) ZAOD (iv) ZABD​

Answer 1

✌️❣️_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_

Given P is equidistant from points A and B

PA=PB .....(1)

and Q is equidistant from points A and B

QA=QB .....(2)

In △PAQ and △PBQ

AP=BP from (1)

AQ=BQ from (2)

PQ=PQ (common)

So, △PAQ≅△PBQ (SSS congruence)

Hence ∠APQ=∠BPQ by CPCT

In △PAC and △PBC

AP=BP from (1)

∠APC=∠BPC from (3)

PC=PC (common)

△PAC≅△PBC (SAS congruence)

∴AC=BC by CPCT

and ∠ACP=∠BCP by CPCT ....(4)

Since, AB is a line segment,

∠ACP+∠BCP=180

∘

Thus, AC=BC and ∠ACP=∠BCP=90

∘

∴,PQ is perpendicular bisector of AB.

Hence proved.

Tusen Takk ✌️❣️