Question

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.
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Help me its urgent . my class test is going on ​

Answer 1

$\huge\bold{\displaystyle\rm\red{Answer}}$

Given : AB/PQ = AC/PR= AD/PM

To prove : ∆ABC ~ ∆PQR

Construction : Produce AD to E so that AD = DE and Join CE.

Produce PM to N so that PM =MN and Join RN.

Proof :

In ∆ADB and ∆EDC,

DB = CD (AD is the median)

AD = DE ( by construction)

∠ADB = ∠CDE ( vertically opposite angles)

∆ADB ≅ ∆EDC ( SAS congruency)

AB = EC ( by cpct) -------1

In ∆PMQ and ∆NMR,

QM = MP (PM is the median)

PM = MN (by construction)

∠PMQ = ∠NMR ( vertically opposite angles)

∆PMQ ≅ ∆NMR ( SAS congruency)

PQ = RN ( by cpct) -------2

In ∆ACE and ∆PRN,

AB/PQ = AC/PR = AD/PM

CE/RN = AC/PR = AD/PM ------[ from 1 and 2]

CE/RN = AC/PR = 2AD/2PM

CE/RN = AC/PR = AE/PN

∆ACE ~ ∆PRN ( SSS similarity)

∠2 = ∠4 (cpst)

Similarly, ∠1 = ∠3 --------3

∠2 + ∠1 = ∠4 + ∠3

∠BAC = ∠QPR ---------4

In ∆ABC and ∆PQR,

AB/PQ = AC/PR (given)

∠BAC = ∠QPR ( from 4)

∆ABC ~ ∆PQR (SAS similarity)