Question

Question 14 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

Class 10 - Math - Triangles Page 141

Answer 1

Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that AB/PQ = AC/PR = AD/PM


To Prove: ΔABC ~ ΔPQR


Construction: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.


Proof: In ΔABD and ΔCDE, we have

AD = DE  [By Construction]

BD = DC [∴ AP is the median]

and, ∠ADB = ∠CDE [Vertically opp. angles]

∴ ΔABD ≅ ΔCDE [By SAS criterion of congruence]

⇒ AB = CE [CPCT] ...(i)

Also, in ΔPQM and ΔMNR, we have

PM = MN [By Construction]

QM = MR [∴ PM is the median]

and, ∠PMQ = ∠NMR [Vertically opposite angles]

∴ ΔPQM = ΔMNR [By SAS criterion of congruence]

⇒ PQ = RN [CPCT] ...(ii)

Now, AB/PQ = AC/PR = AD/PM

⇒ CE/RN = AC/PR = AD/PM ...[From (i) and(ii)]

⇒ CE/RN = AC/PR = 2AD/2PM

⇒ CE/RN = AC/PR = AE/PN [∴ 2AD = AE and 2PM = PN]

∴ ΔACE ~ ΔPRN [By SSS similarity criterion]

Therefore, ∠2 = ∠4

Similarly, ∠1 = ∠3

∴ ∠1 + ∠2 = ∠3 + ∠4

⇒ ∠A = ∠P ...(iii)

Now, In ΔABC and ΔPQR, we have

AB/PQ = AC/PR (Given)

∠A = ∠P [From (iii)]

∴ ΔABC ~ ΔPQR [By SAS similarity criterion]

Answer 2

Step-by-step explanation:

★ 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.

★ CONCEPT USED :

(i) Median divides a side of a triangle equally into two equal halves.

(ii) There are three criteria for similarity of triangle

1. SSS similarity criterion (Side - Side - Side)
2. AA similarity criterion (Angle - Angle)
3. SAS similarity criterion (Side - Angle - Side)

(iii) Corresponding angles of similar triangles are equal.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

★ GIVEN :

AD and PM are median of ∆ABC and ∆PQR respectively such that $\sf{\frac{AB}{PQ} \: = \: \frac{AC}{PR} \: = \: \frac{AD}{PM} }$

★ TO PROVE :

ΔABC ∼ ΔPQR

★ CONSTRUCTION :

• Produce AD to E such that AD = DE
• Produce PM to N such that PM = MN
• Join BE and EC
• Join QN and NR

★ PROOF :

(i) In ∆ABD and ∆ECD , we have

AD = DE [By Construction]

∠BDA = ∠CDE [Vertically Opposite Angles]

BD = CD [ ∴ D is the midpoint of BC]

∴ ∆ABD ≅ ∆ECD [By SAS congruency criterion]

AB = EC — — — (i) [CPCT]

(ii) In ∆PQM and ∆NRM , we have

PM = NM [By Construction]

∠PQM = ∠NRM [Vertically Opposite Angles]

QM = RM [ ∴ M is the midpoint of QR]

∴ ∆PQM ≅ ∆NRM [By SAS congruency criterion]

PQ = NR — — — (ii) [CPCT]

(iii) In ∆ADC and ∆EDB , we have

AD = DE [By Construction]

∠ADC = ∠EDB [Vertically Opposite Angles]

DC = DB [ ∴ D is the midpoint of BC]

∴ ∆ADC ≅ ∆EDB [By SAS congruency criterion]

AC = EB — — — (iii) [CPCT]

(iv) In ∆PMR and ∆NMQ , we have

PM = NM [By Construction]

∠PMR = ∠NMQ [Vertically Opposite Angles]

MR = MQ [ ∴ M is the midpoint of QR]

∴ ∆PMR ≅ ∆NMQ [By SAS congruency criterion]

PR = NQ — — — (iv) [CPCT]

$\rm{\frac{AB}{PQ} \: = \: \frac{AC}{PR} \: = \: \frac{AD}{PM}} \: \: \: \: \: \: \rm{[Given]}$

$\implies \: \rm{\frac{EC}{NR} \: = \: \frac{AC}{PR} \: = \: \frac{AD}{PM}} \: \: \: \: \: \rm[Using \: (i) \: and \: (ii) ]$

$\implies \rm{\frac{EC}{NR} \: = \: \frac{AC}{PR} \: = \: \frac{2AD}{2PM}}$

$\implies \: \rm{\frac{EC}{NR} \: = \: \frac{AC}{PR} \: = \: \frac{AE}{PN}} \: \: \: {[2AD \: = \: AE]} {[2PM \: = \: PN]}$

$\rm{\triangle{ACE} \: \sim\: \triangle{PNR}} \: \: \rm{[SSS \: similarity \: criterion]}$

∴ ∠2 = ∠4 — — — (v) [Corresponding angles of similar triangles are equal]

$\rm{\frac{AB}{PQ} \: = \: \frac{AC}{PR} \: = \: \frac{AD}{PM}} \: \: \: \: \: \: \rm{[Given]}$

$\implies \: \rm{\frac{AB}{PQ} \: = \: \frac{EB}{NQ} \: = \: \frac{AD}{PM}} \: \: \: \: \: \rm[Using \: (iii) \: and \: (iv) ]$

$\implies \: \rm{\frac{AB}{PQ} \: = \: \frac{EB}{NQ} \: = \: \frac{2AD}{2PM}}$

$\implies \: \rm{\frac{AB}{PQ} \: = \: \frac{EB}{NQ} \: = \: \frac{AE}{PN}} \: \: \: \rm{[2AD \: = \: AE]} \: \rm{[2PM \: = \: PN]}$

$\rm{\triangle{ABE} \: \sim\: \triangle{PQN}} \: \: \rm{[SSS \: similarity \: criterion]}$

∴ ∠1 = ∠3 — — — (vi) [Corresponding angles of similar triangles are equal]

Adding (v) and (vi). We get ,

∠2 + ∠1 = ∠4 + ∠3

∠1 + ∠2 = ∠3 + ∠4

∠BAC = QPR — — — (vii)

Now , in ∆ABC and ∆PQR , we have

$\rm{ \frac{AB}{PQ} \: = \: \frac{AC}{PR} \: \: [Given]}$

∠BAC = QPR — — — (From vii)

∴ ∆ABC ∼∆PQR [By SAS similarity criterion]

Hence Proved.