Question

Q . +=
slanding Shapes-ll (Quadrilaterals)
le 15 in Fig. 16.18, bisectors of ZB and LD of quadrilateral ABCD meet CD and AB
16.15
in APBC, we have
1
(2 ABC + ZADC).
2
ZP +24+ZC = 180°
Mon
А.
B
(1)
3
4
1
ZP+=ZB+ ZC = 180°
2
In AQAD, we have
ZQ+ZA+Z1 = 180°
1
ZQ+ZA+ZD=180°
2
2
(ii)
P​

Answer 1

Answer:

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