Hi guys!!
Can someone help me out with this proof, please?
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Step-by-step explanation:
same theorem is applied for 3 equations
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To Prove: AM = ½ [Perimeter of ∆ABC]
Given: AM and AN is tangent.
Proof: I hope you remember that length of tangents are equal due to theorem.
→ AM = AN __(1)
→ BM = BP __(2)
→ CN = CP __(3)
Perimeter of ∆ABC = (AB + BC + CA)
→ AB + BP + PC + CA
→ AB + BM + CN + CA [Due to (2) & (3)]
→ AM + AN
→ AM + AM [Due to (1)]
→ 2 AM = Perimeter of ∆ABC
→ AM = ½[P of ∆ABC]
Q.E.D
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