AD & PM are medians. AB/PQ=BC/QR=AD/PM. Prove that ∆ABC ~ ∆PQR.
Answers
It is given that ΔABC ~ ΔPQR
We know that the corresponding sides of similar triangles are in proportion.∴ AB/PQ = AC/PR = BC/QR ...(i)
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R …(ii)
Since AD and PM are medians, they will divide their opposite sides.∴ BD = BC/2 and QM = QR/2 ...(iii)
From equations (i) and (iii), we get
AB/PQ = BD/QM ...(iv)
In ΔABD and ΔPQM,
∠B = ∠Q [Using equation (ii)]
AB/PQ = BD/QM [Using equation (iv)]
∴ ΔABD ~ ΔPQM (By SAS similarity criterion)⇒ AB/PQ = BD/QM = AD/PM
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Answer:
A lim is the value that a function "approaches" as the input "approaches" some value. Limits are essential to calculus and are used to define continuity, derivatives, and integrals.