Question

Given that ∆ABC ~ ∆PQR , CM and RN are respectively the medians of ∆ABC and ∆PQR , Prove that

(1) ∆AMC ~ ∆PNR

(2) CM / RN = AB / PQ

(3) ∆CMB ~ ∆RNQ ​

Answer 1

$\huge\mathbb\red{Given~:-}$

∆ABC ~ ∆PQR

CM and RN are the medians of ∆ABC , ∆PQR respectively

$\huge\mathbb\blue{Required~To~Proof~:-}$

(1) ∆AMC ~ ∆PNR

(2) CM / RN = AB / PQ

(3) ∆CMB ~ ∆RNQ

$\huge\mathbb\pink{Proof~:-}$

∆ABC ~ ∆PQR

angle A = angle P

angle B = angle Q

angle C = angle R

_______and______

AB / PQ = BC / QR = AC / PR

CM and RN are medians

so , AB = 2 AM and PQ = 2 PN

AB / PQ = AC / PR

=> 2 AM / 2 PN = AC / PR

=> AM / PN = AC / PR

In ∆AMC , ∆PNR

The corresponding sides ratio is equal

AM / PN = AC / PR

The corresponding angles are equal

angle A = angle P

By SAS similarity

★ ∆AMC ~ ∆PNR → (1)

Then ,

The corresponding sides ratio is equal

AM / PN = CM / RN = AC / PR

AM / PN = CM / RN

★ 1/2 AB / 1/2 PQ = CM / RN

★ CM / RN = AB / PQ → (2)

We know that

AB / PQ = BC / QR

so ,

2 MB / 2 NQ = BC / QR

MB / NQ = BC / QR

In ∆ CMB and ∆ RNQ

MB = BC [ Corresponding sides ]

NQ = QR [ included angle ]

angle B = angle Q [ included angle ]

By SAS similarity

★ ∆CMB ~ ∆RNQ → (3)

$\huge{\boxed{\mathbb\pink{\fcolorbox{red}{purple}{ItzCUTEstar}}}}$