In triangle ABC , seg XY || side BC , If M and N are the midpoints of dev AY and dev AC respectively . prove that, 1.Triangle AXM~triangle ABN 2.seg XM || dev BN
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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
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