Question

In given figures two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of △PQR.
Show that:
(i) △ABM ≅△PQN
(ii) △ABC ≅△PQR​

Answer 1

$\large \sf \underline{Solution - }$

Here it is given that,

• AB = PQ
• BC = QR
• AM = PN

To prove :

• (i) △ABM ≅△PQN
• (ii) △ABC ≅△PQR

Proof :

(i) In △ABM and △PQN

• AB = PQ$\:\: \: \:\: \:$ [Given]
• AM = PN$\:\: \: \:\: \:$ [Given]
• BC = QR$\:\: \: \:\: \:$ [Given]

As BC and QR are not included in the triangles.

So,

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

• BM = QN

Therefore,

➢△ABM ≅△PQN

• by SSS congurance rule.

(ii) In △ABC and△PQR

• AB = PQ$\:\: \: \:\: \:$ [Given]
• ∠ B = ∠Q$\:\: \: \:\: \:$ [By CPCT]
• BC = QR$\:\: \: \:\: \:$ [Given]

Therefore,

➢△ABC ≅△PQR

• by SAS congurance rule.