Ratio, Similarity and Symmetry

For comparing two quantities, we consider their —

To determine the ratio of two quantities, the two quantities must be measured in —

If $a:b = x:y$ and $c:d = x:y$, then —

If $a:b = b:a$, then —

By inversendo, if $a:b = x:y$, then —

By alternendo, if $a:b = x:y$, then —

By cross-multiplication, $a:b = c:d$ implies —

By componendo, if $a:b = x:y$, then —

By dividendo, if $a:b = x:y$, then —

Componendo and dividendo together give —

If $3:5 = x:15$, then $x = ?$

If $4:7 = 8:y$, then $y = ?$

Applying inversendo on $3:4 = 6:8$ gives —

Applying alternendo on $2:5 = 6:15$ gives —

If two triangles have equal heights, their areas are proportional to their —

If two triangles have equal bases, their areas are proportional to their —

The area of a triangle equals —

Triangles $ABC$ and $DEF$ have equal heights $h$, and bases $BC=a$, $EF=d$. Then $\triangle ABC : \triangle DEF = ?$

Triangles with equal bases $b$ and heights $h$, $k$ have area ratio —

Two triangles have equal bases and heights in ratio $3:5$. Their areas are in ratio —

Two triangles have equal heights and bases $4$ cm and $9$ cm. The ratio of their areas is —

A line parallel to one side of a triangle intersects the other two sides —

In $\triangle ABC$, $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$. Then —

The proof of Theorem 28 uses the fact that $\triangle BDE$ and $\triangle DEC$ have —

Corollary 1 of Theorem 28 gives —

Corollary 2 of Theorem 28 says: a line through the midpoint of one side parallel to another side —

In $\triangle ABC$, $DE \parallel BC$. If $AD = 4$, $DB = 2$, $AE = 6$, then $EC = ?$

In $\triangle ABC$, $DE \parallel BC$. If $AD = 3$, $DB = 5$, $AE = 6$, then $EC = ?$

In $\triangle ABC$, $DE \parallel BC$, $AD:DB = 2:3$. Then $AE:EC = ?$

In $\triangle ABC$, $DE \parallel BC$, $AB = 9$, $AD = 3$, $AC = 12$. Then $AE = ?$

The heights of $\triangle ADE$ and $\triangle BDE$ (with the common vertex pattern of the proof) are —

If a line segment divides two sides of a triangle (or their produced sections) proportionally, it is —

Theorem 29 is the converse of —

In $\triangle ABC$, if $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ where $D \in AB$, $E \in AC$, then —

In the proof of Theorem 29, $\triangle BDE = \triangle DEC$ because they are on the same base $DE$ and —

If $D$ on $AB$ with $AD = 6$, $DB = 4$, and $E$ on $AC$ with $AE = 9$, $EC = 6$, then $DE$ is —

The internal bisector of an angle of a triangle divides the opposite side in the ratio of —

In $\triangle ABC$, $AD$ bisects $\angle A$ and meets $BC$ at $D$. Then —

In the proof of Theorem 30, the auxiliary line $CE$ is drawn —

In the proof, $\angle AEC = \angle BAD$ because they are —

In the proof, $\angle ACE = \angle CAD$ because they are —

In the proof, $AC = AE$ because $\triangle AEC$ is —

In $\triangle ABC$, $AB = 6$, $AC = 4$, $BC = 10$. $AD$ bisects $\angle A$, meets $BC$ at $D$. Then $BD = ?$

In $\triangle ABC$, $AB = 8$, $AC = 6$, $BC = 14$. If $AD$ bisects $\angle A$, $BD = ?$

In $\triangle ABC$, $AB = 5$, $AC = 10$, $BC = 12$. If $AD$ bisects $\angle A$, $BD = ?$

If a side of a triangle is divided internally in the ratio of the other two sides, the segment to the opposite vertex —

In Theorem 31, the construction draws $CE$ parallel to —

In the proof of Theorem 31, $BA:AE = BD:DC$ follows from —

In Theorem 31, $AE = AC$ is established from —

Bisectors of two base angles of a triangle meet the opposite sides at $X$ and $Y$. If $XY \parallel$ base, the triangle is —

If two transversals cut a set of parallel lines, the matching intercepted segments are —

The diagonals of a trapezium intersect each other —

The line joining the midpoints of the oblique sides of a trapezium is —

In $\triangle ABC$, medians $AD$, $BE$ meet at $G$. A line through $G$ parallel to $DE$ meets $AC$ at $F$. Then $AC = ?$

In $\triangle ABC$, $X$ on $BC$, $O$ on $AX$. Then $\triangle AOB : \triangle AOC = ?$

Congruence is a special case of —

Two congruent figures are —

Two similar figures are —

Polygons with equal numbers of sides whose angles are sequentially equal are —

Two polygons are similar when matching angles are equal AND matching sides are —

A rectangle and a square are equiangular but generally —

Two equiangular triangles are always —

Two similar triangles are always —

If two equiangular triangles have one matching pair of sides equal, they are —

If two triangles are equiangular, their matching sides are —

For $\triangle ABC \sim \triangle DEF$ with $\angle A = \angle D$, $\angle B = \angle E$, $\angle C = \angle F$ —

In the proof of Theorem 32, $\triangle APQ \cong \triangle DEF$ by —

After $\triangle APQ \cong \triangle DEF$, we get $\angle APQ = \angle ABC$, hence —

The relation $\dfrac{AB}{AP} = \dfrac{AC}{AQ}$ in the proof follows from —

If three pairs of corresponding sides of two triangles are proportional, the corresponding angles are —

In the proof of Theorem 33, $\triangle APQ \cong \triangle DEF$ by —

In the proof, $\triangle ABC$ and $\triangle APQ$ are equiangular because —

Theorem 33 is the converse of —

If one angle of a triangle equals an angle of another and the sides adjacent to those angles are proportional, the triangles are —

In the proof of Theorem 34, $\triangle APQ \cong \triangle DEF$ by —

After establishing $\dfrac{AB}{AP} = \dfrac{AC}{AQ}$, we conclude —

Theorem 34 is the analogue, for similarity, of —

The ratio of the areas of two similar triangles equals the ratio of the squares of —

If $\triangle ABC \sim \triangle DEF$, then $\dfrac{\triangle ABC}{\triangle DEF} = ?$

$\triangle ABC \sim \triangle DEF$ with $BC = 4$, $EF = 8$. The ratio of areas is —

Two similar triangles have areas in ratio $9:25$. The ratio of matching sides is —

$\triangle ABC \sim \triangle DEF$ with $AB = 6$, $DE = 9$, area of $\triangle ABC = 24$. Area of $\triangle DEF = ?$

In the proof of Theorem 35, $\dfrac{h}{p} = ?$

Two similar triangles with matching side ratio $2:3$ have area ratio —

If areas of two similar triangles are $50$ and $200$, the ratio of matching sides is —

If $X$ lies between $A$ and $B$ with $AX:XB = m:n$, then $X$ divides $AB$ —

Construction 12 divides a segment in a given internal ratio using a parallel line whose existence relies on —

In Construction 12, on ray $AX$ we cut $AE = m$ and $EC = n$. Drawing $ED \parallel CB$ gives —

Example 1: a $7$ cm segment is divided in the ratio $3:2$. The two parts measure —

To divide a $10$ cm segment in ratio $3:7$, the parts are —

In $\triangle ABC$, $DE \parallel BC$ with $D \in AB$, $E \in AC$. Then $\triangle ABC$ and $\triangle ADE$ are —

With $DE \parallel BC$ in $\triangle ABC$, the correct relation is —

With $DE \parallel BC$, the area ratio is —

In Exercise 14.2 Q1, the true statements are —

In $\triangle ABC$, $PQ \parallel BC$ with $P \in AB$, $Q \in AC$. The correct relation is —

If each of two triangles is similar to a third triangle, then they are —

Two right-angled triangles with one pair of acute angles equal are —

The perpendicular from the right-angle vertex of a right triangle splits it into two triangles each similar to —

In Exercise 14.2 Q11, $\triangle ABC : \triangle DEF$ when $\angle A = \angle D$ equals —

In Q12, two similar triangles have $BC = 3$ cm, $EF = 8$ cm, and area $\triangle ABC = 3$ sq cm. Then area $\triangle DEF = ?$

In a parallelogram $ABCD$, a line through $A$ meets $BC$ at $M$ and $DC$ at $N$. Then $BM \times DN = ?$

If two equiangular triangles $ABC$ and $DEF$ have heights $AM$, $DN$ from matching vertices, then —

A figure has line symmetry if it can be folded about a line so that the two parts —

The line of symmetry is also called —