Question

CD and RS are respectively the medians of ABC and PQR as shown below.

Answer 1

Solution :-

Since CD and RS are medians of ∆ABC and ∆PQR .

So,

→ AD = DB = (1/2)AB

→ PS = SQ = (1/2)PQ

given that, ∆ABC ~ ∆PQR,

→ AB/PQ = BC/QR = CA/RP { Corresponding parts of similar ∆'s are in proportion . }

→ AB/PQ = CA/RP

→ 2AD/2PS = CA/RP

→ AD/PS = CA/RP -------- Eqn.(1)

also,

→ ∠BAC = ∠QPR { Corresponding angles of similar ∆'s are equal . }

→ ∠DAC = ∠SPR -------- Eqn.(2)

then, In ∆ADC and ∆PSR from Eqn.(1) and Eqn.(2) we get,

→ ∆ADC ~ ∆PSR { By SAS similarity. } (Proved.)

So,

→ AD = PS { Corresponding sides of similar ∆'s are equal. }

Or,

= DB = SQ -------- Eqn.(3)

also,

→ DC = SR { Corresponding sides of similar ∆'s are equal. }

----------- Eqn.(4)

now, in ∆CDB and ∆RSQ we have,

→ DB = SQ { From Eqn.(3) }

→ DC = SR { From Eqn.(4) }

→ BC = QR { ∆ABC ~ ∆PQR. }

hence,

→ ∆CDB ~ ∆RSQ { By SSS similarity. } (Proved)

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