Question

In the figure, ∠TMA ≡∠IAM and ∠TAM ≡ ∠IMA. P is the midpoint of MI and N is the midpoint of AI. Prove that ΔPIN ~ ΔATM

Answer 1

Proof:

S.No.	Statements	Reasons
1.	∠TMA ≡ ∠IAM ∠TAM = ∠IMA	given
2.	∠ATM = ∠IMA	Remaining angle By angle sum property
3.	IP = PM ⇒ `"IP"/"PM"` = 1 In = NA ⇒ `"IN"/"NA"` = 1	P is the midpoint of IM and N is the midpoint of IA
4.	`"IP"/"PM" = "IN"/"NA"`	By 3
5.	PN || MA	By 4
6.	∠IPN = ∠IMN ∠INP = ∠IAM	By 5
7.	In ΔPIN and ΔATM (i) ∠IPN = ∠TAM (ii) ∠INP = ∠TAM (iii) ∠ATM = ∠PIN	By 1, 2 and 6
8.	ΔPIN ∼ ΔATM	By AAA criteria