Question

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

If ABC ~ PQR then, prove that:

(a) ADC ~ PSR (2)

(b) CDB ~ RSQ by using ADC ~ PSR​

Answer 1

Solution →

Given, Two similar Triangles ABC and PQR

CD ! AB and. RS ! PQ

To prove : ∆ ADC ~ ∆ PSR

Proof : CA / RP = AB / PQ

→ CA / RP = 2 AD / 2PS

( since D & S are medians of AB and PQ )

Now, in ∆ ACD & ∆ PRS,

CA / RP = AD / PS

and < A = < P ( therefore ∆ ABC ~ ∆ PQR )

< ADC = < PSR = 90⁰

Thus, by A - A similarity criterion,

•™ ∆ ADC ~ ∆ PSR ( Hence proved )

♦ Given : ∆ ADC ~ ∆ PSR

♦ To prove : CD / RS = AB / PQ

♦ Proof : Given that ∆ ABC & ∆ PQR are similar.

so, < A = < P ... (1)

( corresponding angles of similar Triangles are same )

CA / RP = AB / PQ = CB / RQ .... (2)

Now, in ∆ CAD and ∆ RPS ,

• < A = < P ( given )

< CDA = < RSP = 90 ⁰

By the A-A similar criterion,

∆ CDA ~ ∆ RSP

CA / Rp = CD / RS

from the equation ,

CD / RS = AB / PQ

♪ Hence proved,

:)