Triangles with the same base and height have the same area

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What to assume they know

Half base times height, and that any of the three sides may serve as base — Why the area of a triangle is half base times height
That a rectangle's diagonals bisect each other
Congruence, and that congruent figures have equal areas
Parallel lines, and that the distance between two parallels is the same everywhere
Reflection in a line, and that a point and its image are the same distance from the line
That the shortest route between two points is the straight segment joining them

What they should be able to do

State the condition under which two triangles must have equal areas, and distinguish it from congruence
Show that the four triangles cut from a rectangle by its two diagonals have equal areas, naming the base-height pair used for each comparison
Prove that a line from a vertex to the midpoint of the opposite side splits a triangle into two of equal area
Given a family of triangles on one base with the apex on a fixed parallel line, say which has the greatest and which the least area, and justify the answer
Say which member of that family has the least perimeter, and which has the greatest, and explain why one of these questions has no answer
Reproduce the reflection argument for the minimum-perimeter triangle, naming the congruence that makes the two path lengths equal
Confirm that the minimising apex lies on the perpendicular bisector of the base
Compute an area or a fraction of an area using the equal-halves result rather than by measuring anything new
Apply the reflection argument to a route problem stated outside geometry

Where it usually goes wrong

"Equal area means congruent." The chapter opens this subsection by saying the four triangles are not congruent, and then proves they are equal. Getting students to hold those two ideas apart is most of the work of the explanation.
"The tallest-looking triangle in the family has the most area." They all have the same area. The apparent height a student sees is a slant length, not the perpendicular distance.
"If the area does not change, nothing changes." The perimeter changes continuously, and has a minimum. Two measures, one picture.
"A median divides a triangle into two congruent halves." Equal in area, not congruent — except in the isosceles case, where it happens to be both and therefore teaches the wrong lesson if it is met first.
"Any line from a vertex halves the triangle." Only the one to the midpoint. Show a near-miss line and the unequal halves it leaves.
"The minimum-perimeter apex is found by trying positions and measuring." It is found by reflecting, and the reflection converts a bent path into a straight one. That conversion is the idea worth remembering.
"The shortest way to the river is straight down." Only if you are not going on to the tank afterwards. The second destination is what makes it a reflection problem.
"Both parts of question (ii) have answers." The least perimeter exists; the greatest does not. A question can be well posed and still have no answer, and saying so is honest mathematics rather than a dodge.
"The tick marks on the Part II p.157 figures are decoration." They mark the equal distances that make the perpendicular-bisector conclusion work, and they are the chapter's way of pointing at the answer to its own Math Talk.

Questions to check understanding

Decide whether two triangles in a given figure have equal areas, and name the base-height pair that settles it
Show that a stated segment halves a triangle's area, or that it does not
Given the area of one piece of a divided triangle, find the area of the whole
Find what fraction of a triangle is taken by the triangle on two of its midpoints
Given a family of triangles on one base between parallels, answer both the area question and the perimeter question, stating which has no answer
Locate the point on a line that minimises a two-leg route, and justify it by reflection
Given a trapezium, construct a triangle of equal area
Explain why two triangles can have the same area without being congruent — the reasoning item this subsection is built to support

Examples worth working on the board

Items marked printed are stated or worked on the page; items marked not in the book are an added argument or arithmetic on the chapter's inputs and must not be presented as something the chapter states.

Four triangles from a rectangle's diagonals (Part II p.155, two figures). Rectangle ABCD with both diagonals drawn, crossing at O; the four triangles are numbered 1 on the left, 2 at the top, 3 on the right and 4 at the bottom. The question is whether the four have equal areas. Printed: the chapter first disposes of the easy answer — the four are not four congruent triangles — and then compares two adjacent ones, 1 and 2. Taking OD and OB as the bases, both have the same altitude, drawn in the second figure from A to the diagonal DB with its foot marked X. And OB = OD because a rectangle's diagonals bisect each other. So triangles 1 and 2 are equal, and the same argument run round the figure makes all four equal. Note: the shared altitude is the one step a student will not think of, because it is an altitude of both triangles at once.
The general statement (Part II p.155, bottom, with a marginal figure). Printed: joining a vertex to the midpoint of the opposite side divides a triangle into two triangles of equal area. The marginal figure shows a triangle split into pieces 1 and 2 by such a segment, with a printed caption saying the two agree in both base and height. Not in the book: the two halves are almost never congruent — their third sides differ — and that is exactly why the result needs the area argument rather than a congruence argument.
The whole family on one base (Part II p.156, top figure and question). A line l is drawn parallel to BC, and several triangles are drawn on the base BC with their apexes at different points of l. Printed questions: (i) which of these has the greatest area and which the least, and (ii) which has the greatest perimeter and which the least. Printed: the chapter says it will only show how to find the least perimeter and leaves the rest as exercises. Not in the book answers to (i): every triangle in the family has the same area, because they share the base BC and their heights are all the constant distance between the two parallels — so every triangle in the family is at once a largest and a smallest. Question (i) has an answer; what it lacks is a unique one. Keep that distinct from the perimeter half of (ii), where no maximum exists at all — this brief builds a whole misconception bullet on exactly that difference, and using the same words ("no greatest and no least") for both cases hands the student the wrong reason for (i). That is why the chapter could safely leave it as an exercise, and saying so is more useful than giving a name.
The perimeter question (Part II p.156, second figure). Printed: intuition suggests the least perimeter comes from the apex on the perpendicular bisector of BC, and the chapter then asks how that could be justified; the figure shows ∆ABC with A above the midpoint, the right angle marked at the foot, and tick marks showing the two halves of BC equal. Printed: since BC is common to all the triangles, only the sum of the other two sides matters. Not in the book for the other half of question (ii): there is no greatest perimeter. Slide the apex far enough along l and the two slanted sides grow without any bound, so the question as printed has one answer and one non-answer, and a student should be told which is which.
The mirror (Part II pp.156–157). Printed: treat the line l as a plane mirror and reflect everything below it; the chapter appeals explicitly to the experimental fact from science that an image sits as far behind the mirror as the object sits in front. B and C reflect to B′ and C′. Printed: because ∆AXB and ∆AXB′ are congruent, AB = AB′, and likewise AC = AC′; therefore the route from B to A to C is exactly as long as the route from B to A to C′. Printed: this holds wherever A sits on l.
Straightening the path (Part II p.157). Printed: so the problem of finding the A that minimises B to A to C is the same as finding the A that minimises B to A to C′ — and that one is easy, because the shortest route from B to C′ is the straight segment BC′; take A where BC′ crosses l. The chapter concludes that this triangle has the least perimeter, and then sets a Math Talk asking the student to settle whether that apex sits on the base's perpendicular bisector.
Answering the Math Talk (not in the book; the chapter poses it and stops). It does. Because BC is parallel to l, B and C are the same distance below l, so C′ is that same distance above it. The segment from B to C′ therefore crosses l exactly halfway across, which is the point directly above the midpoint of BC. The tick marks the chapter draws on its Part II p.157 figures are marking those equal distances, which is the hint that the argument is meant to be finished this way.
An isosceles triangle halved (Part II p.158, Figure it Out 3). ∆SUB is isosceles with S at the apex, U and B at the ends of the base, and E the foot of the perpendicular from S to UB; tick marks mark the two equal sides. The area of ∆SEB is given as 24 sq. units, and the area of ∆SUB is asked for. Not in the book: 48 sq. units, because in an isosceles triangle the perpendicular from the apex lands on the midpoint of the base, so SE is the halving segment of the general statement and the two halves are 24 each. Note that here the two halves are congruent — which is a special case, not the reason.
The midpoint triangle (Part II p.159, Figure it Out 7, marked Try This). The points M and N bisect the two sides XY and XZ, and the question asks what part of ∆XYZ is taken by ∆XMN; tick marks on the figure mark all four half-sides, and the printed hint is to join NY. Not in the book: joining NY gives ∆XNY, which is half of ∆XYZ because YN runs from Y to the midpoint of XZ; then inside ∆XNY the segment NM runs from N to the midpoint of XY, so ∆XMN is half of ∆XNY. The answer is one quarter, reached by applying the halving result twice and measuring nothing.
Gopal's errand (Part II p.159, Figure it Out 8, a Math Talk). The figure shows a river drawn as a band across the upper part of the page, a labelled point for the water tank below it, and a labelled point for the house lower still and to the left. Gopal starts at the house, must reach the river, and must then reach the tank; the shortest such route is wanted, and the student is asked to redraw the map and trace it. Not in the book: reflect the tank across the river's near bank, draw the straight segment from the house to that image, and the point where it meets the bank is where Gopal should go. It is the Part II p.157 argument with the triangle removed, and it is the moment to say that the mirror trick was never about triangles.
A trapezium turned into a triangle (Part II p.170, Figure it Out 8). ZYXW is a trapezium with ZY parallel to WX, A is the midpoint of XY, and B lies out on the line through W and X; the item asks for a proof that the trapezium and ∆ZWB have the same area. Not in the book: the segment ZA extended meets that line at B, and ∆ZYA and ∆BXA are congruent because A bisects XY and the two parallels give equal alternate angles — so swapping the one for the other converts the trapezium into the triangle without changing the total. This item is printed in the trapezium exercise set and Area of a trapezium, derived two different ways lists it as data too; agree the boundary before scripting if the two videos go to different producers.
One more instance, owned elsewhere (Part II p.160, Figure it Out 4). Two blue triangles inside a rectangle, one standing on the top edge and one on the bottom edge, meeting at a common apex inside. Not in the book: their two heights add to the rectangle's height, so together they are exactly half of it whatever the apex position. Any polygon is a pile of triangles owns this figure.

Figures to have open

The rectangle with both diagonals and four numbered regions (Part II p.155), plus the second printed figure showing the shared altitude with its foot at X. Both are needed; the argument lives in the second one.
The marginal halving figure (Part II p.155) with the two pieces tinted differently, so that "equal but not congruent" is visible at a glance.
The family of triangles on one base between two parallels (Part II p.156). Redraw with more apex positions than the printed figure shows, because the explanation wants the slide to be continuous. This is the topic's key visual.
The mirror construction (Part II p.157, three figures). Redraw as one step-by-step figure rather than three stills; the printed version repeats the whole picture at each stage, which an explanation does not need to do.
The Gopal map (Part II p.159, item 8) with the tank's reflected image added. The reflection is not in the printed figure and is the whole answer.
The midpoint triangle (Part II p.159, item 7) with the hint segment NY drawn in a second colour.
The trapezium and ∆ZWB (Part II p.170, item 8) with the two congruent corner triangles tinted, showing the swap.
No photograph is needed. The river in the Part II p.159 figure is drawn as two plain lines with a label, not as artwork.

Where this sits in the book

NCERT Ganita Prakash Class 8, Part II, printed Chapter 7, "Area", §7.1 "Rectangle and Squares". The material sits under the smaller-tier subheading "Some Applications of the Area Formula" on Part II p.155 and then under the subheading "Triangles between Parallel Lines with a Common Base" on Part II pp.156–157. The worked BY calculation at the head of Part II p.155 belongs to Why the area of a triangle is half base times height.
Part II pp.158–159, Figure it Out items 3, 7 and 8. Items 1 and 2 belong to Why the area of a triangle is half base times height; items 4 and 5 are the Śulba-Sūtras transformations, also carried there; item 6, the three identical squares, belongs to Any polygon is a pile of triangles.
Part II p.170, Figure it Out 8, for the trapezium-to-triangle proof. That item is printed inside the trapezium exercise set and is listed by Area of a trapezium, derived two different ways as well.
Part II p.160, Figure it Out 4, referenced only as a forward pointer; owned by Any polygon is a pile of triangles.
Part II p.171, the chapter SUMMARY. Note that the SUMMARY does not carry the equal-base-equal-height result or the halving result, although both are stated in the body. A revision list built from the SUMMARY alone will miss this whole topic.
Backward pointer: Why the area of a triangle is half base times height for half base times height and for the fact that any side may be taken as the base, which is what licenses the base-swapping in section 2.