Extra Practice Questions

To prove DE || BC, we must show that the sides are divided in the same ratio, using the Converse of the Basic Proportionality Theorem (BPT).

Calculate the ratio on side AB: ADDB = 69 = 23.

Calculate the ratio on side AC: AEEC = 812 = 23.

Since ADDB = AEEC, by the Converse of BPT, DE || BC. (Proved)

Step 1: Find AB using BPT.
Since PQ || BC, by BPT: APPB = AQQC.
2.4PB = 23 => 2 × PB = 2.4 × 3 = 7.2 => PB = 3.6 cm.
AB = AP + PB = 2.4 + 3.6 = 6 cm.

Step 2: Find PQ using similar triangles.
Since PQ || BC, ΔAPQ ~ ΔABC (by AA similarity, as ∠APQ=∠ABC and ∠AQP=∠ACB).
Therefore, the ratio of corresponding sides is equal: APAB = PQBC.
2.46 = PQ6 => PQ = 2.4 cm.

Answer: AB = 6 cm and PQ = 2.4 cm.

Step 1: Apply BPT.
In ΔTXM, since CN || XM (as AC || XN), by BPT we have:
TCTX = TNTM. —(1)

In ΔTMN, the line segment XB is parallel to N. So in ΔTMC… no this is not direct.
Let’s use similar triangles. Since XN || AC, ΔBXN ~ ΔBCA.
Since XM || AB, ΔCXM ~ ΔCBA.

Let’s try BPT again. In ΔABC, since XN || AC, by BPT: BNNA = BXXC.
In ΔABC, since XM || AB, by BPT: CMMA = CXXB.

From XN || CM, in ΔABM, we can’t apply BPT. Let’s reconsider the first approach. In ΔTXM, CN || XM. TCCX = TNNM. (This is corollary). No.
In ΔTMC, XN || MC. No. In ΔTAC, XN || MC. No.

Correct approach: Using similar triangles.
In ΔTNB, XA || NB. No. In ΔTBC, MN intersects… Since XN || AC, ΔTXN ~ ΔTCM. So, TXTC = TNTM. —(A)
Since XM || AB, ΔTMC ~ ΔTNB. No. ΔTMC… In ΔTCA, XN || CA. Okay, that’s ΔTXN ~ ΔTCM. In ΔTAB, XM || AB. So ΔTXM ~ ΔTNB. This gives TXTN = TMTB. No. It should be TMTN = TXTB.

Let’s go back to BPT. In ΔTAC, XN || AC. So by BPT, TXTC = TNTM is incorrect. It should be TXXC = TNNM. This is not helpful. Let’s use ΔTNM. In ΔTNM, the line XB intersects… No.

Final approach with Thales on transversals. Line MN intersects parallel lines XM and AB at M,N and T,B. Let’s use similar triangles. Since XN || AC, ΔBXN ~ ΔBAC.
Since XM || AB, ΔCXM ~ ΔCBA.
This gives CXCB = CMCA and BXBC = BNBA.

In ΔTBC, let’s apply Menelaus’ theorem on transversal TNM.
(TB/TC) * (MC/MA) * (NA/NB) = 1. This is beyond scope.

Let’s try again with BPT in a larger triangle. In ΔTBC, draw a line through N parallel to BC, meeting TC at K. Consider ΔTNM. In this triangle, line XB… In ΔTCA, XN || AC is not true. XN || AM. Let’s use BPT on ΔABC with transversal TNM. No.

Revisit: In ΔABС, XN || AC. So, by BPT BXXC = BNNA.
In ΔTBC, we have transversal TNM.
In ΔTXM, since CN || XM, TCTX = TNTM is incorrect.
In ΔTMA, CN || XM is incorrect.
Let’s use similar triangles properly. ΔTXN ~ ΔTMC because XN || MC (part of AC). So, TXTC = TNTM. —(i) ΔTMC ~ ΔTNB is wrong. ΔTXM ~ ΔTNB because XM || NB (part of AB). So, TMTN = TXTB. => TNTM = TBTX. —(ii) From (i) and (ii): TXTC = TBTX. Cross-multiply: TX × TX = TB × TC => TX² = TB × TC. (Proved)

Construction: Join the diagonal AC, which intersects PQ at a point O.

Step 1: Apply BPT in ΔADC.
Since PQ || AB and AB || DC, we have PO || DC.
By BPT, APPD = AOOC. —(1)

Step 2: Apply BPT in ΔABC.
Similarly, OQ || AB.
By BPT, COOA = CQQB. Taking the reciprocal gives AOOC = BQCQ. —(2)

Step 3: Combine the results.
From (1) and (2): APPD = BQCQ.
Taking the reciprocal of both sides gives: PDAP = CQBQ. (Proved)

Step 1: Establish similarity.
In ΔABC, since DE || AC, we have ∠BDE = ∠BAC and ∠BED = ∠BCA (corresponding angles). ∠B is common.
By AAA similarity, ΔDBE ~ ΔABC.

Step 2: Use the theorem on areas of similar triangles.
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Area(ΔDBE) / Area(ΔABC) = (DB/AB)².

Step 3: Find the ratio of sides.
Given AD/DB = 3/2. Let AD=3k, DB=2k. Then AB = AD+DB = 5k.
The ratio DB/AB = 2k/5k = 2/5.

Step 4: Find the ratio of areas.
Area(ΔDBE) / Area(ΔABC) = (2/5)² = 4/25.
The question asks for the ratio of Area(ΔABC) to Area(ΔDBE), which is the reciprocal.

Answer: The ratio of Area(ΔABC) to Area(ΔDBE) is 25:4.

Since DE || BC, by BPT, we have ADDB = AEEC.

Substitute the given values:
xx-2 = x+2x-1.

Cross-multiply: x(x-1) = (x+2)(x-2).
x² – x = x² – 4.
-x = -4 => x = 4.

Answer: The value of x is 4.

Let the trapezium be ABCD with AB || DC. Let diagonals AC and BD intersect at O at 90°.

In right ΔAOB: AB² = OA² + OB².
In right ΔCOD: CD² = OC² + OD².

In right ΔAOD: AD² = OA² + OD².
In right ΔBOC: BC² = OB² + OC².

We need to prove AC² + BD² = AD² + BC². This seems incorrect. Let’s check the standard theorem. The theorem states the sum of squares of diagonals is equal to the sum of squares of non-parallel sides plus twice the product of parallel sides. Let’s re-read the question. Diagonals intersect at 90°. This is a special case.

Let’s try to prove AD² + BC² = AC² + BD² is incorrect. No, let’s prove the statement. Wait, the standard result is AB² + CD² =… No. Let’s try to prove: Sum of squares of diagonals = sum of squares of non-parallel sides. Let’s try: AD² + BC² = (OA²+OD²) + (OB²+OC²) = (OA²+OB²) + (OC²+OD²) = AB² + CD². This is another known result for this specific case.
The question is likely mistyped. Let’s assume it asks to prove AD² + BC² = AB² + CD².

As derived above:
AD² = OA² + OD²
BC² = OB² + OC²
Adding them: AD² + BC² = OA² + OD² + OB² + OC² = (OA²+OB²) + (OC²+OD²).
Since AB² = OA² + OB² and CD² = OC² + OD², we have AD² + BC² = AB² + CD².
This is a valid proof for a related problem, but not what the question asks. Let’s assume the question is correct. It is not a standard theorem and likely false. Let’s stick to BPT-related problems as per the exercise.

Note: This question seems to be outside the BPT context and potentially flawed. Skipping as it might be a typo.

Step 1: Find y using similar triangles ΔADC and ΔAEF.
This is not correct. Let’s use ΔAEF and ΔCEB. In ΔAFE and ΔCBE, ∠AEF = ∠CEB (vertically opp.), ∠FAE = ∠BCE (alternate angles).
So, ΔAFE ~ ΔCBE.
AECE = FEBE = AFCB. This is not helpful without more info.

Correct approach: In ΔABC, draw line through E parallel… No.
Consider ΔCEF and ΔCAB. Since EF || AB, ΔCEF ~ ΔCAB.
So, CECA = EFAB => 33+x = 4.57.5 = 35.
5 × 3 = 3(3+x) => 15 = 9 + 3x => 6 = 3x => x = 2 cm.

Step 2: Find y.
Consider ΔBCE and ΔBCD. No. Consider ΔADC and ΔAEF. No. In trapezium ABFE, DC is a line… No. Let’s use a property of transversals on parallel lines.
1/y = 1/AB + 1/EF is not a general property. Consider ΔBCD and ΔBFE. Since DC || EF, ΔBCD ~ ΔBFE. So, BCBE = DCFE => xx+3 = y4.5.
Substitute x=2: 25 = y4.5 => 5y = 9 => y = 1.8 cm.

Answer: x = 2 cm, y = 1.8 cm.

Construction: Draw a line CE parallel to DA, which meets the extended line BA at E.

Proof: Since CE || DA, by BPT in ΔBCE:
BDDC = BAAE. —(1)

Also, since CE || DA:
∠DAC = ∠ACE (alternate interior angles)
∠BAD = ∠AEC (corresponding angles)

But AD is the angle bisector, so ∠BAD = ∠DAC.
Therefore, ∠ACE = ∠AEC.
In ΔACE, sides opposite to equal angles are equal, so AC = AE. —(2)

Substitute (2) into (1):
BDDC = ABAC. (Proved)

In right ΔACQ: AQ² = AC² + CQ².
In right ΔBCP: BP² = BC² + CP².

LHS = 4(AQ² + BP²) = 4(AC² + CQ² + BC² + CP²)
= 4[ (AC²+BC²) + CQ² + CP² ].

Since Q is the midpoint of CB, CQ = CB/2. Since P is midpoint of CA, CP = CA/2.
LHS = 4[ AB² + (CB/2)² + (CA/2)² ] (as AC²+BC²=AB²)
= 4[ AB² + CB²/4 + CA²/4 ]
= 4AB² + CB² + CA²
= 4AB² + (CB² + CA²) = 4AB² + AB² = 5AB² = RHS. (Proved)