Category Archives: Proofs in plane geometry

Midpoint Theorem

In the figure, AB is parallel to PRQ, T and P are the midpoints of CQ and CA respectively and ART is a straight line. Prove that TR : RA = 1  : 2.

Solution

Point Q is the midpoint of BC (midpoint theorem)

QB : QC                 QC : TQ

   1 :  1                      2  :  1

      x2                          x1

   2  :  2                     2  :  1

QB  :  TQ

  2   :   1

∠TRQ = ∠TAB (corresponding RQ//AB)

TQ/QB = TR/RA

1/2 = TR/RA

Therefore TR : RA = 1 : 2

∠TQR = ∠TBA (corresponding ∠s, RQ//AB)

Therefore triangle TRQ is similar to triangle TAB

Problems involving properties of similar triangles

In the diagram, ABCD is a rectangle and PC is perpendicualr to BD. Prove that

(i) Triangle ABD is similar to triangle PDC

(ii) AB2 = BD x PD

(i) 

∠ABD = ∠PDC (alternate ∠s, AB//OC)

∠DPC = ∠BAD = 90°(given)

Therefore by Angle Angle similarity, triangle ABD is similar to triangle PDC.

(ii)

AB/PD = BD/DC

AB x DC = BD x PD

Since, AB = DC

Therefore AB= BD x PD

Proving using angle properties of circles

In the figure, BCE and ADE are straight lines. Given that ∠BAD = ∠ACD, prove that 

(i)CD bisects ∠ACE,

(ii) BD = AD

Solution:

(i) Let ∠BAD and ∠ACD be x

∠DCB = 180° – x (∠s in the opposite segments)

∠ECB = 180° – (180° – x)

∠ECD = x

Since , ∠ECD = ∠ACD = x

Therefore CD bisects ∠ACE

 

(ii) ∠DBA = x (∠s in the same segment)

since, ∠DBA = ∠BAD = x (base ∠s of isoceles triangle)

Triangle DBA is an isoceles triangle

therefore BD = DA