Revision: Similarity Geometry Maths 2 SSC (English Medium) 10th Standard Maharashtra State Board

In similar triangles, the angles opposite to proportional sides are the corresponding angles, and so, they are equal. 

• ∠A = ∠P

• ∠B = ∠Q

• ∠C = ∠R

In similar triangles, the sides opposite to equal angles are said to be the 
corresponding sides. 

ΔABC ∼ ΔPQR

\[\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}\]

Two triangles are similar if

• Their corresponding angles are equal, and
• Their corresponding sides are proportional.
• Symbolically:
  ΔABC ∼ ΔPQR (read as “ABC is similar to PQR”).

Statement:
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

To Prove:
$DE\parallel BC$

$Proof:$

1. Assume a line through point D parallel to BC meets AC at F.

2. By BPT, \[\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AF}}{\mathrm{FC}}\]

3. Given, \[\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\]

4. Hence,\[\frac{AF}{FC}=\frac{AE}{EC}\]

5. ⇒ Points E and F coincide.

   Therefore,

   $DE\parallel BC$

Statement:
If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.

To Prove:
\[\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\]

Proof:

1. A line parallel to a side of a triangle forms equal corresponding angles.

2. Hence, the two triangles formed are similar (AAA similarity).

3. In similar triangles, corresponding sides are proportional.

Therefore, the line divides the two sides in the same ratio.

\[\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\]

Statement:
When two triangles are similar, the ratio of the areas of those triangles is equal to the ratio of the squares of their corresponding sides.

\[\frac{\mathrm{BC}^{2}}{\mathrm{QR}^{2}}=\frac{\mathrm{AB}^{2}}{\mathrm{PQ}^{2}}=\frac{\mathrm{AC}^{2}}{\mathrm{PR}^{2}}\]

• corresponding altitudes
• corresponding medians
• corresponding angle bisectors

Statement:
The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

\[\frac{AE}{EB}=\frac{CA}{CB}\]

Statement:
The ratio of the intercepts made on a transversal by three parallel lines is equal to the ratio of the corresponding intercepts made on any other transversal by the same parallel lines.

\[\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{PQ}}{\mathrm{QR}}\]

• AA / AAA → two angles equal

• SAS → included angle equal + sides proportional

• SSS → all sides proportional

• Ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.

• Areas of triangles with equal heights are proportional to their corresponding bases.

• Areas of triangles with equal bases are proportional to their corresponding heights.