Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 8 · Chapter companion part
Write valid congruence proofs and distinguish a theorem from its converse; justify midpoint, perpendicular-foot and angle-bisector applications.
Class 8 English Version: part B of the chapter 8 companion. Eight teaching sections, 30 written tasks and a notebook challenge. Selected companion coverage, not complete textbook coverage; use all five parts. Independent proofs, compass work, drawings and nets require guardian or teacher review.
Open this topic in Skills Lab →
A proof starts with the givens and the desired conclusion. State a construction such as joining AC, then write each equality with its reason. Congruent triangles have matching vertices in matching order. ASA means two angles and their included side; SAS means two sides and their included angle; SSS means all three sides. Never cite a rule whose required data have not been established.
In parallelogram ABCD, join AC. AB∥CD gives angle BAC= DCA; BC∥AD gives BCA= DAC; AC is common. Thus triangles BAC and DCA are congruent by ASA. Hence AB=CD and BC=DA. Matching remaining angles gives B=D; the split equalities at A, C also give A=C. The diagonal therefore divides the figure into two congruent triangles.
Let diagonals meet at O. In triangles AOB and COD, opposite sides AB=CD and alternate angles at A, C and B, D give ASA congruence, so AO=OC, BO=OD. Conversely, if both diagonals bisect, triangles AOB, COD have equal half-side pairs and equal vertical angle: SAS. Matching alternate angles prove AB∥CD. A second triangle pair proves AD∥BC. Do not assume parallelism in the converse.
For rectangle ABCD compare triangles ABD and DCA: AB=DC, AD=DA, and included angles BAD=CDA=90°. SAS gives BD=CA. Mutual bisection follows because a rectangle is a parallelogram. Printed source page 137 labels this side-angle-side step ASA and mismatches triangle order; the applicable rule is SAS with correspondence ABD↔DCA.
In rhombus ABCD, diagonals bisect at O. Triangles AOB and COB have AB=CB, AO=CO and BO common, so SSS makes AOB=COB. These adjacent angles on straight AC add to 180°, hence each is 90°. A square inherits both rhombus and rectangle properties. Conversely, mutual bisection first gives a parallelogram; perpendicular diagonals give a rhombus and equal diagonals give a rectangle, together a square.
Join consecutive side midpoints of a rectangle. By the triangle midpoint theorem, each small segment is parallel to a rectangle diagonal and half its length. Equal rectangle diagonals make all four small sides equal, so the midpoint figure is a rhombus. In an equilateral triangle, joining midpoints gives segments parallel to the third sides and half their lengths; each smaller triangle has the corresponding 60° angles.
For a parallelogram let O be its diagonal crossing. A half-turn about O maps A to C and B to D. Orthogonal projection onto line BD then maps the foot M from A to the foot N from C. Thus M, N have midpoint O, as do A, C; the diagonals of AMCN bisect, so it is a parallelogram when nondegenerate. If a perpendicular diagonal makes both feet coincide at O, the proposed four-vertex figure degenerates; state this exception.
One opposite side pair that is both equal and parallel suffices: a diagonal gives SAS, then equal alternate angles establish the other parallel pair. A parallelogram with equal diagonals is a rectangle: compare triangles ABC and BAD by SSS, giving adjacent equal supplementary angles and hence 90°. With equal adjacent sides it is a rhombus. Printed page 143 question 17 calls BACD the vertex order while naming AC, BD diagonals; use explicit cyclic ABCD for the intended argument and record the naming inconsistency.
For rectangle ABCD, triangles ABD and DCA match A↔D, B↔C, D↔A. The known two sides and included right angle establish SAS, so BD=CA. Correct correspondence is part of the proof.
Printed page 137 uses ASA beside a two-side/included-angle argument. The companion uses SAS and matching triangle order.
If the two perpendicular feet coincide, four distinct vertices do not exist. Such a special case is explained instead of marked as a valid parallelogram.
Compare triangle ABD with triangle DCA.
| Statement | Reason |
|---|---|
| AB=DC | Opposite rectangle sides |
| AD=DA | Common side |
| BAD=CDA=90° | Rectangle right angles |
| ABD≅DCA | SAS: included angle between the two known sides |
| BD=CA | Corresponding sides |
The source labels this step ASA. Its supplied data justify SAS; triangle order must match A↔D, B↔C, D↔A.
A converse starts with the old conclusion and must be proved independently.
| Given in a simple nondegenerate quadrilateral | Conclusion | Reason route |
|---|---|---|
| Both diagonals bisect | Parallelogram | SAS on half-diagonal triangles |
| One opposite side pair both equal and parallel | Parallelogram | SAS across one diagonal |
| Parallelogram and equal diagonals | Rectangle | SSS gives equal supplementary adjacent angles |
| Parallelogram and perpendicular diagonals | Rhombus | SAS on right half-triangles gives adjacent sides equal |
| Equal perpendicular diagonals with mutual bisection | Square | Rectangle and rhombus together |
Equal diagonals alone and perpendicular diagonals alone are not sufficient. Degenerate four-point arrangements are excluded.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original labelled mathematical model, not copied source artwork. Verify stated conditions rather than judging shape from appearance.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original labelled mathematical model, not copied source artwork. Verify stated conditions rather than judging shape from appearance.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original labelled mathematical model, not copied source artwork. Verify stated conditions rather than judging shape from appearance.
Show your working on paper. Check your answer, then choose Next.
3
SSS means side-side-side.
8
Opposite sides AB, CD are equal.
5
BC=DA.
6
Mutual bisection gives equal halves.
9
Each half is 18/2.
17
SAS proves the diagonal lengths equal.
90
Equal adjacent angles on a straight line each equal 180/2.
5
Each midpoint segment is half a rectangle diagonal.
20
Four midpoint sides of 5 cm give 20.
6
The midpoint theorem gives half the third side.
6
3−(−3)=6; O is their midpoint.
90
The equal-diagonal converse makes it a rectangle.
4
The conditions make a square.
70
180−42−68=70; state the triangle angle-sum reason.
107
They sum to 180°.
Write in your own words. A guardian or teacher should review this response.
AB∥CD and BC∥AD give two pairs of alternate angles about AC. The shared AC is their included side. ASA proves triangles BAC≅DCA with the correct correspondence.
Write both angle equalities and the common-side reason.
Write in your own words. A guardian or teacher should review this response.
With O their crossing, AB=CD and alternate angles BAO=DCO, ABO=CDO give triangles ABO≅CDO by ASA. Matching sides yield AO=CO and BO=DO.
First establish the opposite-side equality; do not assume bisection.
Write in your own words. A guardian or teacher should review this response.
AO=CO, BO=DO and vertical angles give AOB≅COD by SAS, so corresponding alternate angles imply AB∥CD. Similarly AOD≅COB gives AD∥CB.
Use two pairs and name SAS; no parallel premise may be assumed.
Write in your own words. A guardian or teacher should review this response.
AB=DC, AD=DA, included BAD=CDA=90°. Hence ABD≅DCA by SAS and BD=CA. The source ASA label is corrected.
Match triangle vertices and show the included angle.
Write in your own words. A guardian or teacher should review this response.
Bisection gives AO=CO. AB=CB and BO common give AOB≅COB by SSS. Equal angles AOB, COB are supplementary, so each is 90°.
Both equality and straight-line sum are needed.
Write in your own words. A guardian or teacher should review this response.
Bisection gives a parallelogram by the converse. Equal diagonals give a rectangle. The perpendicular-bisection triangles have equal corresponding adjacent sides, making a rhombus. A figure both rectangle and rhombus is a square.
Explain each implication; perpendicularity alone does not suffice.
Write in your own words. A guardian or teacher should review this response.
The triangle midpoint theorem makes two midpoint segments half AC and the other two half BD. A rectangle has AC=BD, so all four new sides are equal. Their parallelism also follows from the midpoint theorem.
Identify the parent triangles and all four new sides.
Write in your own words. A guardian or teacher should review this response.
The midpoint theorem in ABC gives DF∥BC and DF=BC/2. B, D, F, E form a simple quadrilateral, so their interior-angle sum is 360°. In this equilateral setting the angles are 60°,120°,120°,60°.
Keep the new cyclic order B, D, F, E explicit.
Write in your own words. A guardian or teacher should review this response.
Half-turn symmetry about O sends A↔C and the line BD to itself, so M↔N and both diagonals AC, MN have midpoint O. Their bisection proves AMCN a parallelogram if the four vertices are distinct. If AC⊥BD, M=N=O and the construction degenerates.
Include the exception rather than claiming a four-vertex figure always exists.
Write in your own words. A guardian or teacher should review this response.
If AB=CD and AB∥CD, join AC. Alternate BAC=DCA, shared AC and AB=CD give BAC≅DCA by SAS. Matching BCA=DAC proves BC∥AD; both opposite pairs are parallel.
Use the same pair for both the equal and parallel premises.
Write in your own words. A guardian or teacher should review this response.
In ABC and BAD, AB is common, BC=AD and AC=BD. SSS gives ABC=BAD. These adjacent parallelogram angles sum to 180°, so each is 90° and the figure is a rectangle.
The equality premise alone for an arbitrary quadrilateral would fail.
Write in your own words. A guardian or teacher should review this response.
Because BO is a median, AO=OC. Given BO=OD, both diagonals AC and BD bisect at O. Apply the proved converse.
A median is defined by the midpoint of the opposite side, not by perpendicularity.
Write in your own words. A guardian or teacher should review this response.
Equal adjacent sides together with equal opposite sides make all four equal, hence a rhombus. Its diagonals meet at 90° by the SSS and supplementary-angle proof.
Do not assume equal diagonals for a nonsquare rhombus.
Write in your own words. A guardian or teacher should review this response.
In cyclic BACD, AC and BD would be sides. Use intended cyclic ABCD explicitly. A+B=180° gives AD∥BC by the converse of the same-side interior-angle rule; B+C=180° similarly gives AB∥CD. Thus ABCD is a parallelogram and AB=CD, AD=BC. Unequal diagonals do not prevent this.
State the naming correction, prove both parallel pairs from the supplied angle sums, and only then apply the opposite-side theorem.
Write in your own words. A guardian or teacher should review this response.
Opposite interior bisectors have the same direction, so AB∥CD; BC lies on CQ and AD on PA, which are parallel. Thus ABCD is a parallelogram when nondegenerate. No right-angle condition was supplied.
Record where B, D lie and exclude a collapsed segment.
Write three two-column proofs and one corrected counterexample.
Keep your work on paper or in a local file. This page does not upload it or automatically grade the project.
This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 8, printed pages 133–151. It does not replace the complete chapter or its practical teaching.
Official book catalogue and source →
Teacher-led lessons, schedules and prices for this outline are upcoming. You can use any starter practice above now without booking.
Solo practice works without an account. Sign in to save a practice record to your learner profile.
Exact lesson links and related activities are labelled separately. Return-to-course links retain your selected learning path.
Play this exact lesson in Skills Lab →