Transversal And Mid-point Theorem

In Δ ABC, since D is the mid-point of BC and as FD || AB,

Then by the theorem:-

Through the mid-point of any side, if a line segment is drawn parallel to second side, then it will bisect the third side and the line segment intercepted by the two sides of the triangle is equal to half of the second side.

⇒ F is the mid-point of AC

By using the above theorem, we can also prove that E is the midpoint of AB as DE || AC.

Now as E and F have been proved to be the midpoint of AB and AC respectively, by applying the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get,

In ΔABC, Let F and G be the midpoint of AB and AC respectively.

So, in ΔAFG, and ,

⇒ D and E are the midpoint of AF and AG respectively.

By applying the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get,

and DE || FG ……… (1)

Also, by using above theorem in ΔABC, we get,

and FG || BC ……… (2)

From equations (1) and (2), we get

and DE || BC

In ΔPQR, as X and Z are the midpoint of QR and PQ respectively.

So by applying the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get,

……… (1)

Also, in ΔSZX since ZX || PY and P is the midpoint of SZ,

By applying theorem:-

Through the mid-point of any side, if a line segment is drawn parallel to second side, then it will bisect the third side and the line segment intercepted by the two sides of the triangle is equal to half of the second side.

……… (2)

From equations (1) and (2), we get

In ΔACD, as E and H are the midpoints of AC and CD respectively.

So by applying the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get,

and EH || AD ……… (1)

Also, by using above theorem in ΔABD, as F and G are midpoints of AB and BD respectively,

and FG || AD ……… (2)

From equations (1) and (2) we see that, EH = FG and EH || AD.

As both the above conditions are sufficient for a parallelogram,

⇒ EFGH is a parallelogram

In ΔACD, as E and H are the midpoints of AC and CD respectively.

So by applying the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get,

and EH || AD ……… (1)

Also, by using above theorem in ΔABD, as F and G are midpoints of AB and BD respectively,

and FG || AD ……… (2)

From equations (1) and (2) we see that, EH = FG and EH || AD.

Similarly for ΔBCD, we have G and H as midpoints of BD and CD respectively, by applying above theorem, we get,

and GH || BC ……… (3)

And also in ΔABC,

and EF || BC ……… (4)

From equations (3) and (4) we see that, GH = EF and GH || EF.

As ABCD is a rectangle, the length of diagonals are equal

⇒ AD = BC

⇒ HG = GF = FE = EH

But, we also see that since ABCD is a rectangle not square BD is not equal to CD,

∴ ∠ DCB is not equal to ∠ DBC.

Hence, the angle between the sides in EFGH cannot be 90° and as a result EFGH is not a square figure but a rhombus.

In ΔACD, as E and H are the midpoints of AC and CD respectively.

So by applying the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get,

and EH || AD ……… (1)

Also, by using above theorem in ΔABD, as F and G are midpoints of AB and BD respectively,

and FG || AD ……… (2)

From equations (1) and (2) we see that, EH = FG and EH || AD.

Similarly for ΔBCD, we have G and H as midpoints of BD and CD respectively, by applying above theorem, we get,

and GH || BC ……… (3)

And also in ΔABC,

and EF || BC ……… (4)

From equations (3) and (4) we see that, GH = EF and GH || EF.

As ABCD is a square, the length of diagonals are equal

⇒ AD = BC

⇒ HG = GF = FE = EH

⇒ AE = AF, this means ∠AEF = ∠AFE = ∠ABC = ∠ACB = 45°

∴ We can easily prove that angle between the sides of the figure FGHE is equal to 90°

Hence, Quadrilateral FGHE is a square.

In ΔACD, as E and H are the midpoints of AC and CD respectively.

So by applying the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get,

and EH || AD ……… (1)

Also, by using above theorem in ΔABD, as F and G are midpoints of AB and BD respectively,

and FG || AD ……… (2)

From equations (1) and (2) we see that, EH = FG and EH || AD.

Similarly for ΔBCD, we have G and H as midpoints of BD and CD respectively, by applying above theorem, we get,

and GH || BC ……… (3)

And also in ΔABC,

and EF || BC ……… (4)

From equations (3) and (4) we see that, GH = EF and GH || EF.

So, EHGF is a parallelogram and as IH lie on EH and HJ lie on HG.

⇒ IHJO is also a parallelogram.

As we know that, the angle subtended by the diagonals in a triangle is equal to 90°, and also in a parallelogram, the opposite angles are equal.

⇒ ∠EHG = ∠IOJ = 90°

Hence, quadrilateral HGFE is a rectangle

In ΔBDC, as Q and P are the midpoints of BD and CD respectively.

So by applying the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get,

and PQ || BC ……… (1)

Similarly, applying above theorem on ΔABC where D and E are midpoints of AB and AC respectively

and DE || BC ……… (2)

From equations (1) and (2),

⇒ PQ = DE and PQ||DE, ……… (3)

⇒ PQDE is a parallelogram.

Also in ΔBDE, Q is the midpoint to BD and QH||DE as PQDE is a parallelogram,

……… (4)

From equations (3) and (4), it proves that:

⇒ BE and PQ bisect each other

We extend the perpendicular till F such that it intersects the line BC at point F.

As BG is the angle bisector of ∠ABC,

⇒ ∠ABD = ∠FBD = θ

Since AD is perpendicular to BG, so ∠BAD = 90 – θ

In ΔABF, as sum of all sides of triangle is equal to 180°

⇒ ∠ABF + ∠BAF + ∠BFA = 180

⇒ 2θ + (90 – θ) + ∠BFA = 180

⇒ ∠BFA = 90 – θ

We know that sides opposite to equal angles in a triangle are equal and as ∠BFA = ∠BAD.

⇒ BA = BF

In ΔBAD and ΔBFD,

∠ABD = ∠FBD (given)

∠ADB = ∠FDB = 90° (given)

BA = BF (proved above)

⇒ ΔBAD ≅ ΔBFD by RHS congruency.

⇒ AD = DF, that is, D is the midpoint of AF.

Now, In ΔACF, as D is midpoint of AF and DE || FC, then by applying theorem:-

Through the mid-point of any side, if a line segment is drawn parallel to second side, then it will bisect the third side and the line segment intercepted by the two sides of the triangle is equal to half of the second side, we get,

⇒ E is the midpoint of AC

∴ AE = EC, hence proved.

Since it is given that AD is the median to side BC,

⇒ D is midpoint to BC

Also, it is given that AD||CT||BR, so by applying theorem:-

Through the mid-point of any side, if a line segment is drawn parallel to second side, then it will bisect the third side and the line segment intercepted by the two sides of the triangle is equal to half of the second side.

⇒ TC = RB ……… (1)

As , it can also be written as,

Using equation (1) in above equation,

As E and F are the midpoints of sides AC and BD respectively, therefore by the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get,

⇒ EF || CD

Now, we have extended EF to H, so in ΔACD, since E is midpoint of AC and EH||CD, so by applying theorem:-

Through the mid-point of any side, if a line segment is drawn parallel to second side, then it will bisect the third side and the line segment intercepted by the two sides of the triangle is equal to half of the second side.

Similarly, applying above theorem in ΔADB as F is midpoint of BD and FH||AB,

As FE = FH – EH,

So by substituting above values, we get,

As AR, BS, CT are the minimum distances of the line PQ from the points A, B and C, this can be only achieved when AB||PQ and AR, CT and BS are perpendicular to it.

This makes ARTC and CTSB as a parallelogram.

⇒ CT = AR from ARTC parallelogram,

And CT = BS from CTSB parallelogram.

On adding the above two equations, we get

⇒ AR + BS = 2CT

Since BL||DN||CM and D is midpoint of BC, by using the theorem:-

If three or more parallel straight lines make equal intercepts from a traversal, then they will make equal intercepts from another traversal.

⇒ LN = MN

In ΔDLN and ΔDMN,

DN = DN (common)

∠DNL = ∠DNM = 90° (given)

LN = MN (proved above)

∴ ΔDLN ≅ ΔDMN by SAS congruency.

As a result of it, DL = DM

We have extended PQ till G such that OP||CG,

As O is intersection of Diagonals of a square, therefore O is mid point of AC.

Using the theorem,

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get

……… (1)

As AP is angle bisector of ∠BAO,

⇒ ∠BAP = ∠OAP = θ

⇒ in ΔABQ, ∠AQB = 90 – θ

∠GQC = ∠AQB = 90 – θ (Opposite angles)

Similarly in ΔACG, as ∠ACG = 90° ,

⇒ ∠AGC = 90 – θ

As in ΔGQC, ∠AGC = ∠GQC = 90 – θ and as sides opposite to equal angles in a triangle are equal

⇒ CQ = CG ……… (2)

From equations (1) and (2),

From the given question, we can see that,

Since, S is the midpoint of PR,

Since AB||DC and E and F are midpoints of AD and BC,

By using the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get

⇒ AB||CD||EF

Also,

In Δ ADG, AE = ED and EF||DG, therefore F is mid point of AG.

Similarly, In Δ BFC, BD = DC and BF||DG, therefore G is mid point of FC.

⇒ AF = 3.5 cm

By using the theorem,

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get

⇒ X and Y are them id points of DF and DE respectively.

In ΔABC, by using above theorem,

Similarly applying it in ΔDFE,

From above two equations, we get

AF = AB + BF,

We can see that ΔBEF ≅ ΔCED by AAA congruency

⇒ BF = CD and CD = AB as ABCD is a parallelogram.

⇒ AF = 2AB

We extend the median AD till H such that GE||CH

In Δ BEC, BD = DC and BF||DG, therefore D is midpoint of BC,

By applying theorem:-

Through the mid-point of any side, if a line segment is drawn parallel to second side, then it will bisect the third side and the line segment intercepted by the two sides of the triangle is equal to half of the second side.

Also, as E is midpoint of AC,

From above two equations, we get

∴ length of FC is 2 cm

If R, Q and P are the midpoints of AB, AC and BC, and by applying the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get

and PQ||AR

Also, and PR || AQ

Since the above conditions are sufficient for a parallelogram, therefore ARPQ is a parallelogram.

Perimeter of parallelogram ARPQ = (2×AR) + (2×AQ)

⇒ Perimeter = AB + AC = 30 + 21 = 51 cm

In ΔABD, as X is midpoint of AD and P is midpoint of AB, we apply the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get

and PX parallel to BD

⇒ BD = 2 × PX = 10 cm

Similarly, In ΔBDC, as Y is midpoint of DC and Q is midpoint of BC, we apply the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get

and QY is parallel to BD

⇒ QY = 5 cm

In ΔABC, as F is midpoint of AB and E is midpoint of AC, we apply the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get

and EF||BC

Similarly, In ΔOBC, as P is midpoint of BG and Q is midpoint of GC, we apply the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get

and PQ||BC

⇒ BC = 2× PQ

⇒ BC = 2 × 3

⇒ BC = 6 cm

∴ The length of BC is 6 cm.

In ΔABC, as F is midpoint of AB and D is midpoint of BC, we apply the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get

and FD|| AC

Also, it is given that E is the midpoint of AC

From above two equations, we get, AE = FD

In ΔFOD and ΔEOA,

∠FOD = ∠EOA (Opposite angles)

∠FDO = ∠EAO (Alternate Interior angles as FD || AE)

Also, ∠OFD = ∠OEA (Alternate Interior angles as FD || AE)

⇒ ΔFOD ≅ ΔEOA

∴ AO = OD

∴ The length of AO is 3 cm.