Question

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

Answer 1

Solution :-

In ∆AXY and ∆ABC, we have,

→ ∠AXY = ∠ABC (Since, Seg XY ll side BC, so, corresponding angles .)

→ ∠XAY = ∠BAC (Common.)

So,

→ ∆AXY ~ ∆ABC (By AA similarity.)

then,

→ AX/AB = AY/AC (By CPCT.) ------------ Eqn.(1)

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

So,

→ AY = 2AM

→ AC = 2AN

putting these values in Eqn.(1) we get,

→ AX / AB = 2AM/2AN

→ AX/AB = AM/AN -------------- Eqn.(2)

now, in ∆AXM and ∆ABN, we have,

→ AX/AB = AM/AN {from Eqn.(2)}

and,

→ ∠XAM = ∠BAN (Common.)

so,

→ ∆AXM ~ ∆ABN (By SAS similarity.)

then,

→ ∠AXM = ∠ABN (By CPCT.)

therefore, we can conclude that,

→ seg XM ll seg BN . (Since corresponding angles are equal , therefore, lines will be parallel .)

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