Question

In A ABC, seg XY || side BC. If M and N are the
midpoints of seg AY and seg AC respectively. Prove that
(a) A AXM ~ AABN
(b) seg XM || seg BN.​

Answer 1

Answer :-

from image in ΔAXY and ΔABC we have,

→ ∠AXY = ∠ABC (since XY ∣∣ BC so, corresponding angles.)

→ ∠XAY = ∠BAC (Common angle.)

so,

→ ΔAXY ~ ΔABC (By AA similarity.)

then,

→ AX/AB = AY/AC (By CPCT)

also, we have given that, Point M and N are the midpoints of seg AY seg AC respectively .

so,

→ AY = 2AM

→ AC = 2AN

putting this value , we get,

→ AX/AB = 2AM / 2AN

→ AX/AB = AM / AN

therefore, in ΔAXM and ΔABN we have,

→ AX/AB = AM / AN

so,

→ ∠XAM = ∠BAN (Common angle.)

then,

→ ΔAXM ~ ΔABN (By SAS similarity.)

therefore,

→ ∠AXM = ∠ABN (By CPCT.)

hence, we can conclude that

→ seg XM ∣∣ seg BN (Corresponding angles are equal only if lines are parallel .)

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