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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
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Solution
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 |
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