Question

if ad and pm are alttitude of triangle abc and triangle pqr respectively where triangle abc similar triangle pqr then prove that ab/pq =ad /pm

Answer 1

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

Answer:

ab/pq = ad/pm

Hence proved.

Step-by-step explanation:

SAS criteria:

• If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are said to be similar triangles.

 Here we consider a=A, b=B, c=C, d=D, p=P, q=Q, r=R, m=M

The two triangles ABC and PQR  are similar triangles.

              ΔABC ~ ΔPQR

  by SAS property we can write

      ∠ABC = ∠PQR      (CORRESPONDING ANGLES) ---------(i)

     AB/PQ = BC/QR  (CORRESPONDING SIDES)

                 = (BC/2)/(QR/2)

 Since D and M are midpoints of BC and QR

    then BC/2 =BD and QR/2 = QM.            

           AB/PQ = BD/QM    ----------------(ii)

  In ΔABD and ΔPQM

           ∠ABD = ∠PQM   from(i)

          AB/PQ = BD/QM  from(ii)

Hence ΔABD ~ ΔPQM are corresponding triangles by SAS criteria.

 ⇒AB/PQ = BD/QM = AD/PM (corresponding sides)

⇒ AB/PQ = AD/PM

      Hence ab/pq = ad/pm

Know more about Geometry theorem:

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