Question

4) If seg AB = seg MJ, seg BC = seg
JR, seg AC = seg MR, then the two
triangles are congruent in the
correspondence.
(A) ABC MRJ
(B) ABC RMJ
(C) ACB MRJ
(D) ACB RMJ​

Answer 1

REF.Image

Given, O is the center of a circle.

AB=AC

OP⊥AB

OQ⊥AC

∠PBA=30

∘

To prove :- BP∥QC

Construction : Join BC, OC and OB.

Proof : AB = AC

∠ACB=∠ABC __ (1)

OC=OB [∵ radius]

∠OCB=∠OBC __ (2)

∴∠ACB=∠OBC=∠ABC−∠OBC [ using (1) - (2)]

⇒∠ACO=∠ABO __ (3)

In △OXC and △OYB

∠OXC=∠OYB[∵OQ⊥AC&OP⊥AB]

∠AOC=∠ABO [using (3)]

OC=OB

∴△OXC=≅△OYB [AAS]

∠QOC=∠POB [CPCT] __ (4)

∴ we proved that segPB∥segQC.

Now in △QOC and △POB

OQ=OB[∵ radius]

∠QOC=∠POB (using (4))

OC=OP(∵ radius)

∴△QOC≅△POB [SAS]

∴OQC=∠OBP [CPCT]

∴QC=BP

ACB MRJ