Question

In ABC, seg XY || side BC. If M and N are the

midpoints of seg AY and seg AC respectively.

Prove that

(a) AXM  ABN

(b) seg XM || seg BN.​

Answer 1

Answer:

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

Step-by-step explanation:

b) seg xm || seg bn

Answer 2

Answer:

attached is your expected answer

hope it helps you!!