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Similar triangles present one of the biggest tools students have to solve many geometry questions. And yet, whenever the opportunity arises, the students often fail to spot the similarity between two triangles. Sometime they spot the similarity but fail to apply the ratios correctly. Today we will start with some familiar cases in which we apply similarity of triangles and then move on to some tricky and difficult cases. We hope that after solving all these problems, you’ll not have any issue in spotting and solving similarity.

In practical scenario, we always prove that two triangles are similar by proving that their corresponding angles are equal. For example, look at the very simple figure below:

If you can prove that $\angle BAC$ = $\angle QPR$ and $\angle ABC$ = $\angle PQR$ then $\triangle ABC$ and $\triangle PQR$ are similar. In which case the ratios of the corresponding sides are equal. The best way to write the ratio of the sides is to write both the triangles in the order of the angle which are equal (for example here $\angle A$ = $\angle P$, $\angle B$ = $\angle Q$ and $\angle C$ = $\angle R$ so write ABC and PQR only) and then write the ratio by picking same corresponding points from the two written triangles. Therefore,

Also, since area = length $\times$ length, the ratio of the areas of the two triangle is square of the ratio of the sides. That means

Specially notice the second figure in the examples above, here $\angle P \neq \angle S$ but equal to $\angle T$. Therefore, writing the triangles in the order of angles which are equal, $\triangle PQR \simeq \triangle TSR$

Now let’s see some problems:

Problem 1: In the diagram given below, $\angle ABD = \angle CDB = \angle PQD = 90^{\circ}$ If AB: CD = 3: 1, the ratio of CD: PQ is (CAT 2003- Leaked)

1. 1: 0.69     2. 1: 0.75     3. 1: 0.72     4. None of these

Problem 2: In the figure, AB // DC. If the areas of $\triangle ABE$ and $\triangle DCE$ are 4 and 9 respectively, find the area of $\triangle BEC$

1. 5     2. 6     3. 6.5     4. 9

Problem 3: In the square ABCD, AB is extended and E is a point on AB such that CE intersects AD at F and BD at G. The length of FG and GC are 3 and 5 units, respectively. What is the length of EF?

1. 4     2. 5     3. $\frac{13}{3}$     4. $\frac{16}{3}$

Problem 4: Consider the triangle ABC shown in the following figure where BC = 12 cm, DB = 9 cm, CD = 6 cm and $\angle BCD = \angle BAC$. What is the ratio of the perimeter of the triangle ADC to that of the triangle BDC? (CAT 2005)
1. $\frac{7}{9}$     2. $\frac{8}{9}$     3. $\frac{6}{9}$     4. $\frac{5}{9}$

Problem 5: Given right triangle ABC, AD = DB, DE is perpendicular to AB, AB = 20 and AC = 12, find the area of the quadrilateral ADEC.

1. 58.5     2. 56     3. 45     4. 37.5

Problem 6: PQRS is a rectangle. Diagonals PS and RQ intersect at point T. A perpendicular TU is drawn on PQ from T. U and R are joined. UR and PS intersect at V and a perpendicular VW is drawn on PQ. Now WR is drawn, intersecting PT at X and a perpendicular XY is drawn on PQ. If PQ = 90 units, find the length of PY.

1. 40     2. 22.5     3. 20     4. 15

If the students notice keenly, this is again a mix of scenario 2 and scenario 4. Let’s solve it step by step. Let’s find out the length of PW first.

$\triangle SRV \simeq \triangle PUV$ therefore, the altitudes will also be in the same ratio as that of the sides. Therefore the ratio of the altitudes = SR: PU = 90: 45 = 2: 1. Let SQ = 12x. Then the heights would be 8x and 4x. Therefore, VW = 4x. Now $\triangle PVW \simeq \triangle PSQ$ Therefore, $\frac{PW}{PQ} = \frac{VW}{SQ}$. Therefore, PW = 30.

Now let’s repeat the same exercise for the next set of triangles.

$\triangle PXW \simeq \triangle SXR$ therefore, the altitudes will also be in the same ratio as that of the sides. Therefore the ratio of the altitudes = SR: PW = 90: 30 = 3: 1. Therefore, XY = 3x. Now $\triangle PXY \simeq \triangle PSQ$ Therefore, $\frac{PY}{PQ} = \frac{XY}{SQ}$. Therefore, PY = 22.5.

Problem 7: In the figure given below, O is a point inside the triangle PQR. Line segments ST, UV, XW are drawn through O, parallel to the sides PR, RQ and PQ respectively. The areas if three triangles SOV, WOU and TOX are 9, 25 and 36 respectively. What is the area of triangle PQR?

1. 148     2. 169     3. 196     4. 200

We can see that the three triangles $\triangle$ SOV, $\triangle$ WUO, $\triangle$ OTX are similar (equal corresponding angles because of parallel lines). Since their areas are in the ratios 9: 25: 36, their sides will be in the ratios 3: 5: 6. So let, SO = 3x, WU = 5x, and OT = 6x. Check the figure below:

Since SOWP is a parallelogram, WP = 3x. For the same reason UR = 6x. Therefore, PR = 3x + 5x + 6x = 14x. Now $\triangle PRQ \simeq \triangle SOV$ and the sides are in the ratio 3x: 14x = 3: 14. Therefore, the areas would be in the ratio, 9: 196. Therefore, the area of triangle PQR = 196

Let’s check out a problem where you cannot see similarity at first sight.

Problem 8: Both right triangle ABC and isosceles triangle BCD, have height 5 cm from base BC = 12 cm. Find area of triangle BEC.

Join A and D, and also drop a perpendicular, DP, from D to BC. Since BDC is an isosceles triangle and AB // DP, AD = BP = 6.

Now, A and D have the same height. Therefore, AD // BC. Therefore, $\triangle ADE \simeq \triangle CBE$ Since $\frac{AD}{BC} = \frac{1}{2}$ therefore, the ratio of the heights also is 1: 2. Therefore, the height of $\triangle BCE = 5 \times \frac{2}{3} = \frac{10}{3}$ Therefore, area $\triangle BCE = \frac{1}{2} \times 12 \times \frac{10}{3} = 20$