6. Triangles

Their corresponding sides are proportional, since PQAB = 1.53 = 12 for all sides.

But their corresponding angles are not equal. All angles of ABCD are 90°, while the angles of PQRS are not 90°.

The quadrilaterals are not similar.

EF || QR if PEEQ = PFFR (converse of the Basic Proportionality Theorem).

In ΔABC, LM || CB. So, by the Basic Proportionality Theorem:

In ΔADC, LN || CD. So:

From (1) and (2):

AMAB = ANAD

In ΔABC, DE || AC. So, by the Basic Proportionality Theorem:

In ΔABE, DF || AE. So:

From (1) and (2):

BFFE = BEEC

In ΔPOQ, DE || OQ. So, by the Basic Proportionality Theorem:

In ΔPOR, DF || OR. So:

From (1) and (2): PEEQ = PFFR.

So, in ΔPQR, the line EF divides PQ and PR in the same ratio.

By the converse of BPT, EF || QR

In ΔPOQ, AB || PQ. So, by the Basic Proportionality Theorem:

In ΔPOR, AC || PR. So:

From (1) and (2): OBBQ = OCCR.

So, in ΔOQR, the line BC divides OQ and OR in the same ratio.

By the converse of BPT, BC || QR

Given: In ΔABC, D is the mid-point of AB and a line DE || BC meets AC at E.

To prove: E is the mid-point of AC.

Proof: Since DE || BC, by Theorem 6.1 (BPT):

But AD = DB (D is the mid-point of AB). So, ADDB = 1.

So, DE bisects the third side AC.

Given: In ΔABC, D and E are the mid-points of AB and AC respectively.

To prove: DE || BC.

Proof: AD = DB and AE = EC.

So, DE divides AB and AC in the same ratio.

By Theorem 6.2 (converse of BPT), DE || BC

Construction: Through O, draw EO || DC (and so EO || AB), meeting AD at E.

In ΔADC, EO || DC. So, by BPT:

In ΔDAB, EO || AB. So:

From (1) and (2): AOOC = BOOD.

Rearranging, AOBO = CODO

Given: AOBO = CODO, i.e., AOCO = BODO ... (1)

Construction: Through O, draw OE || AB, meeting AD at E.

In ΔDAB, EO || AB. So, by BPT:

From (1) and (2): AEED = AOOC.

So, in ΔADC, EO divides AD and AC in the same ratio. By the converse of BPT, EO || DC.

Now, EO || AB and EO || DC. So, AB || DC.

So, ABCD is a trapezium.

DOB is a straight line.

In ΔODC:

Since ΔODC ~ ΔOBA, the corresponding angles are equal.

∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°

In ΔOAB and ΔOCD:

So, ΔOAB ~ ΔOCD (AA similarity criterion).

Corresponding sides of similar triangles are proportional.

OAOC = OBOD

In ΔPQR, ∠1 = ∠2, i.e., ∠PQR = ∠PRQ.

So, PQ = PR (sides opposite to equal angles)

In ΔPQS and ΔTQR:

By SAS similarity criterion, ΔPQS ~ ΔTQR

In ΔRPQ and ΔRTS:

By AA similarity criterion, ΔRPQ ~ ΔRTS

Since ΔABE ≅ ΔACD, their corresponding parts are equal (CPCT).

In ΔADE and ΔABC:

By SAS similarity criterion, ΔADE ~ ΔABC

In ΔABE and ΔCFB:

By AA similarity criterion, ΔABE ~ ΔCFB

Since ΔABC ~ ΔFEG: ∠A = ∠F, ∠B = ∠E and ∠ACB = ∠FGE.

In ΔABC, AB = AC. So, ∠ABC = ∠ACB (angles opposite to equal sides)

In ΔABD and ΔECF:

By AA similarity criterion, ΔABD ~ ΔECF

Given: ABPQ = BCQR = ADPM

AD and PM are medians. So, BD = 12BC and QM = 12QR.

Hence, ΔABD ~ ΔPQM (SSS similarity criterion). So, ∠ABD = ∠PQM, i.e., ∠B = ∠Q.

In ΔABC and ΔPQR:

By SAS similarity criterion, ΔABC ~ ΔPQR

In ΔADC and ΔBAC:

So, ΔADC ~ ΔBAC (AA similarity criterion).

Corresponding sides of similar triangles are proportional.

So, CA² = CB . CD

Given: ABPQ = ACPR = ADPM

Construction: Produce AD to E so that DE = AD. Join BE and CE. Similarly, produce PM to N so that MN = PM. Join QN and RN.

The diagonals of ABEC bisect each other. So, ABEC is a parallelogram. Hence, CE = AB and AE = 2AD.

Similarly, PQNR is a parallelogram. Hence, RN = PQ and PN = 2PM.

Hence, ΔACE ~ ΔPRN (SSS). So, ∠CAE = ∠RPN ... (1)

Similarly, BE = AC and QN = PR. So, ΔABE ~ ΔPQN (SSS). Hence, ∠BAE = ∠QPN ... (2)

Adding (1) and (2): ∠BAC = ∠QPR.

In ΔABC and ΔPQR:

By SAS similarity criterion, ΔABC ~ ΔPQR

At the same time, the sun’s rays make the same angle with the ground. The pole and the tower are vertical.

So, the triangle formed by the pole and its shadow is similar to the triangle formed by the tower and its shadow (AA criterion).

Height of the tower 42 m

Since ΔABC ~ ΔPQR:

AD and PM are medians. So, BD = 12BC and QM = 12QR.

In ΔABD and ΔPQM:

So, ΔABD ~ ΔPQM (SAS similarity criterion).

Hence, ABPQ = ADPM