BCRP is a parallelogram, so CR = BP. BP = PA and ∠AQP = ∠CQR (vertically opposite angles). Combining with CR ‖ BA, ∆APQ ≅ ∆CRQ. (Why?)
Step-by-step solution
Idea: We have one pair of equal sides and one pair of equal angles. The parallel lines CR and BA supply one more pair of equal angles, which completes a congruence test.
- Sides: CR = BP (opposite sides of parallelogram BCRP) and BP = PA, so AP = CR.½ mark
- Angles: ∠AQP = ∠CQR (vertically opposite). Since CR ‖ BA, with AC as transversal, ∠PAQ = ∠RCQ (alternate angles).1 mark
- Two angles and a side not between them are equal, so ∆APQ ≅ ∆CRQ by AAS.½ mark
Answer to write in the exam
AP = BP = CR (P midpoint; BCRP parallelogram)
∠PAQ = ∠RCQ (alternate angles, CR ‖ BA)
∠AQP = ∠CQR (vertically opposite angles)
∴ ∆APQ ≅ ∆CRQ (AAS)
Common mistakes that cost marks
- Writing SAS: the equal angle ∠AQP is not between the equal sides AP and CR.
- Using AQ = QC as a reason; that is what this congruence is used to prove.
- Matching P with C instead of with R.
How this can come in the exam
In the proof above, which pair of angles is equal because CR ‖ BA?
- ∠AQP and ∠CQR
- ∠PAQ and ∠RCQ
- ∠APQ and ∠QCR
- ∠ABC and ∠ACB
Show answer
(B) ∠PAQ and ∠RCQ
∠PAQ and ∠RCQ are alternate angles for the parallel lines BA and CR with transversal AC. ∠AQP = ∠CQR comes from vertically opposite angles, not from the parallels.
Try one yourself
If instead we use ∠APQ = ∠CRQ, which test proves ∆APQ ≅ ∆CRQ?
Show answer
ASA: ∠PAQ = ∠RCQ, AP = CR and ∠APQ = ∠CRQ (alternate angles, transversal PR). The equal sides AP and CR lie between the pairs of equal angles.
More questions like this
- We will later generalise the converse of the Midpoint Theorem by dropping the condition that P is the midpoint of AB and assuming only that PQ ‖ BC. Can you guess the conclusion in this case? (Hint: the name of the more general theorem includes the word proportionality.)
- In ∆ABC, suppose P, Q and R are the midpoints of side AB, AC and BC respectively. Instead of joining the midpoints with each other as we did earlier, draw segments AR, BQ and CP. These are called the medians of ∆ABC. What do you see? Repeat this for several triangles and see that the three medians always seem to be concurrent, meaning they pass through a common point! There is more! Assume for the moment that the medians are always concurrent. Let M be the common point where they meet. Compare the lengths of the two parts CM and MP of the median CP, and do the same for the other two medians. Do you see a pattern?
- Let Y and X be the midpoints of CM and BM respectively. Now we can use the Midpoint Theorem in ∆ABC and in ∆MBC! We get that PQ and XY are parallel to BC and hence to each other. Moreover PQ = XY = BC2. So ∆MPQ ≅ ∆MYX by ASA (How?).
- The three medians of ∆ABC pass through a common point, which divides each of the medians in the ratio 2 : 1, with the longer part connecting to the vertex. This point is called the centroid of ∆ABC.
- Give another proof of the Centroid Theorem using a clever construction that we will describe only partly and ask you to complete. Once again let M be the intersection point of the two medians CP and BQ. We want to show that the intersection point of line AM with side BC is the midpoint of BC. For this, extend AM up to a carefully chosen point S such that BSCM becomes a parallelogram. (Hint: To arrange MC ‖ BS, first note that PM and MC are along the same line. Now where should M be on segment AS? Does your placement of S also prove MB ‖ CS?)
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