急! Properties of Circles 12q18

15a)∠DAB
= ∠ABC (base ∠s, isos.Δ)
= ∠ABC (common)
= ∠ACB (base ∠s, isos.Δ)
∴ ΔABC ~ ΔDAB (A.A.)So ∠ADC = ∠BAC
∴ AB is the tagent to the circle at A. (converse of ∠in alt. segment)
b)Let ∠ABC = ∠ACB = ∠BAD = y , ∠BAC = x , thenx + 2y = 180 (∠ sum of Δ) ... (1)
y - x = 21 (given) ... (2)(1) - 2*(2) :3x = 180 - 42
x = 46∠BAC = 46°

16ai)∠DQA
=∠CAQ + b (ext. ∠ of Δ )
= a + b (equal arcs, equal ∠s)
aii)∠ABP
= a + ∠ABD
= a + b (∠s in the same segment)
= ∠DQA (Proved in ai)
∴ ABRQ is a cyclic quadrilnteral. (converse of ext.∠= int. opp.∠)
b)a = ∠DBC /2 (equal arcs, equal ∠s) = 70/2 = 35°∠BDC
= ∠CAB (∠s in the same segment)
= ∠QAB - ∠QAC
= ∠QAB - a (equal arcs, equal ∠s)
= ∠PRQ - a (ext.∠= int. opp.∠)
= 80° - 35°
= 45°

17a)∠ABQ =∠QPD (ext.∠= int. opp.∠)
while
∠QPD + ∠DCQ = 180° (opp. ∠s supp.) So ∠ABQ + ∠DCQ = 180° ,
∴ AB // DC (int.∠s supp.)
Hence ABCD is a trapezium.
b)Join PQ , ∠XPQ = ∠XCQ (∠s in the same segment) ...... (1)∠AYQ = ∠XCQ + ∠CQY (ext. ∠ of Δ) ...... (2) ∠AYQ
= ∠APQ (∠s in the same segment)
= ∠APX + ∠XPQ ...... (3)
Substitute (1) into (2) :
∠AYQ = ∠XPQ + ∠CQY ...... (4)
Substitute (3) into (4) :
∠APX + ∠XPQ = ∠XPQ + ∠CQY
∠APX = ∠CQY