Question

Here show that ∆ABM is congruent to ∆PQN

b ) ∆ABC is congruent to ∆PQR

GIVEN IS THAT

AM and PN are medians ​

Answer 1

$\mathcal{\gray{\underbrace{\blue{QUESTION:-}}}}$

⚡✨ Two sides AB and BC and median AM of one ∆ABC are respectively equal to sides PQ and QR and median PN of ∆PQR . Show that;

1. ∆ABM $\rm{\cong}$ ∆PQN .
2. ∆ABC $\rm{\cong}$ ∆PQR .

$\mathcal{\gray{\underbrace{\blue{SOLUTION:-}}}}$

✍️ See the attachment figure .

[1] ∆ABM $\rm{\cong}$ ∆PQN .

GIVEN :-

• AB = PQ ------(1)

• BC = QR ------(2)

• AM = PN ------(3)

✍️ Also, AM is the median of ∆ABC .

So,

$\rm\red{\implies\:BM\:=\:CM\:=\:\dfrac{1}{2}\:BC\:}$

✍️ And PN is the median of ∆PQR .

So,

$\rm\red{\implies\:QN\:=\:RN\:=\:\dfrac{1}{2}\:QR\:}$

TO PROVE :-

• $\rm{\triangle{ABM}\:\cong\:\triangle{PQN}\:}$

PROOF :-

Since,

✍️ BC = QR

$\rm{\implies\:\dfrac{1}{2}\:BC\:=\:\dfrac{1}{2}\:QR\:}$

$\rm\red{\implies\:BM\:=\:QN\:}$ -----(4)

✍️ In ∆ABM and ∆PQN,

• AB = PQ ------[from (1)]

• AM = PN ------[from (3)]

• BM = QN ------[from (4)]

$\red\therefore$ ∆ABM $\rm{\cong}$ ∆PQN [SSS congruence rule]

[2] ∆ABC $\rm{\cong}$ ∆PQR

✍️ From part [1] ∆ABM $\rm{\cong}$ ∆PQN .

$\rm\red{\implies\:\angle{B}\:=\:\angle{Q}\:}$ ----(1)

✍️ In ∆ABC and ∆PQR,

• AB = PQ [given]

• ⟨B = ⟨Q [from (1)]

• BC = QR

$\red\therefore$ ∆ABC $\rm{\cong}$ ∆PQR [SAS congruence rule]

Answer 2

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✤ Required Answer:

✒️ GiveN:

• Two sides of a triangle and median is equal to corresponding two sides of another triangle and median.
• The triangles are given as ∆ABC & ∆PQR
• The medians are AM and PN respectively.

✒️ To FinD:

• ∆ABM ≅ ∆PQN
• ∆ABC ≅ ∆PQR

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✤ How to solve?

We will use several criteria's to prove the above triangles by any of the following congruence criterias:

• SSS
• SAS
• AAS
• ASA
• RHS

Here, S means side, A means angle and RHS is Right hypotenuse Side which is used for the congurence of Right angled triangles.

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✤ Solution:

1) We have,

• AB = PQ (Given in Q.)
• AM = PN (Given in Q.)
• BC = PQ (Given in Q.)

Here,

AM is the median of the triangle ABC and hence M is the midpoint of BC. So, BM = MC. Similarly, PN is the median of the triangle PQR and hence N is the midpoint of BC. So, QN = NR.

➝ BC = PQ

➝ 2BM = 2PN

➝ BM = PN

In ∆ABM and ∆PQN,

1. AB = PQ
2. BM = PN
3. BC = PQ

So, ∆ABM ≅ ∆PQN (By SSS congurence)

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2) In Question. 1)

➝ ∆ABM ≅ ∆PQN

So,

➝ ∠ABM = ∠PQN (by CPCT)

In ∆ABC and ∆PQR,

1. AB = PQ
2. ∠ABM = ∠PQN$.$
3. BC = PQ

So, ∆ABC ≅ ∆PQR (By SAS congurence)

❒ Hence, solved !

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