In the figure, angle TMA = angle IAM and angle TAM = angle IMA. find the midpoint of MI and N is the midpoint of AI. Prove that ∆PIN ~ ∆ATM.
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Answer:
Solution:
Given that l1 || l2
∴ In ∆PQS and ∆PRT
∠P is common
∠Q = ∠R [∵ PR is the transversal for l1 and l2 corresponding angles]
∠S = ∠T [∵ corresponding angles]
∴ ∆PQS ~ ∆PRT [∵ By AAA congruency]
In similar triangles, corresponding angles are proportional.
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Step-by-step explanation:
Given that I1 || 12 .. In APQS and APRT ZP is common ZQ = ZR [ PR is the transversal for 11 and 12 corresponding angles] ZS = ZT [: corresponding angles] .: APQS " APRT [ By AAA congruency] In similar triangles, corresponding angles are proportional. Step-by-step explanation:
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