Question

if Ad and PM are the altitude of triangle ABC and triangle pqr respectively where triangle ABC is similar to triangle pqr prove that AB over PQ is equal to AD upon PM

Answer 1

Step-by-step explanation:

Tong ABC ~ PQR :: <B= <Q

also ADand PM are altitudes

:<D=<M ( each 90)

: by AA corollary

; triangleADB~ trianglePMQ

hence AB/PQ

= AD/PM

Answer 2

Hey mate here is your answer

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

Hope it helped you :)