Revision: Geometry Mathematics SSLC (English Medium) Class 10 Tamil Nadu Board of Secondary Education

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”).

A secant is a straight line that passes through the circle and intersects it at two distinct points. 

line AB cuts the circle at points M and N ⇒ AB is a secant.

A tangent is a straight line that touches a circle at exactly one point only, without cutting through it. This single point where the tangent touches the circle is called the point of contact or point of tangency.

Line CD touches the circle at point P only ⇒ CD is a tangent

Point P is the point of contact.

(i) Applying Pythagoras theorem in ΔAMD, we obtain

AM² + MD² = AD² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC²

AM² + (MD + DC)² = AC²

(AM² + MD²) + DC² + 2MD.DC = AC²

AD² + DC² + 2MD.DC = AC² [Using equation (1)]

Using the result, DC = `"BC"/2`, we obtain

`"AD"^2+(("BC")/2)^2 + 2"MD".(("BC")/2) = "AC"^2`

`"AD"^2+(("BC")/2)^2 + "MC" xx "BC" = "AC"^2`

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AB² = AM² + MB²

= (AD² − DM²) + MB²

= (AD² − DM²) + (BD − MD)²

= AD² − DM² + BD² + MD² − 2BD × MD

= AD² + BD² − 2BD × MD

= `"AD"^2+(("BC")/2)^2 - 2(("BC")/2) xx "MD"`

= `"AD"^2 + ("BC"/2)^2 - "BC" xx "MD"`

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AM² + MB² = AB² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC² … (2)

Adding equations (1) and (2), we obtain

2AM² + MB² + MC² = AB² + AC²

2AM² + (BD − DM)² + (MD + DC)² = AB² + AC²

2AM²+BD² + DM² − 2BD.DM + MD² + DC² + 2MD.DC = AB² + AC²

2AM² + 2MD² + BD² + DC² + 2MD (− BD + DC)

= AB² + AC²

= `2("AM"^2 + "MD"^2) + (("BC")/2)^2 + (("BC")/2)^2 + 2"MD" ((-"BC")/2 + ("BC")/2) = "AB"^2 + "AC"^2`

`"2AD"^2 + ("BC"^2)/2 = "AB"^2 + "AC"^2`

Applying Pythagoras theorem in ΔAMD, we obtain

AM² + MD² = AD² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC²

AM² + (MD + DC)² = AC²

(AM² + MD²) + DC² + 2MD.DC = AC²

AD² + DC² + 2MD.DC = AC² [Using equation (1)]

Using the result, DC = `"BC"/2`, we obtain

`"AD"^2+(("BC")/2)^2 + 2"MD".(("BC")/2) = "AC"^2`

`"AD"^2+(("BC")/2)^2 + "MC" xx "BC" = "AC"^2`

Draw perpendicular BD from the vertex B to the side AC. A – D

In right-angled ΔABC

seg BD ⊥ hypotenuse AC.

∴ By the similarity in right-angled triangles

ΔABC ~ ΔADB ~ ΔBDC

Now, ΔABC ~ ΔADB

∴ `"AB"/"AD" = "AC"/"AB"`   ...(c.s.s.t)

∴ AB² = AC × AD   ...(1)

Also, ΔABC ~ ΔBDC

∴ `"BC"/"DC" = "AC"/"BC"`   ...(c.s.s.t)

∴ BC² = AC × DC   ...(2)

From (1) and (2),

AB² + BC² = AC × AD + AC × DC

= AC × (AD + DC)

= AC × AC   ...(A – D – C)

∴ AB² + BC² = AC²

i.e., AC² = AB² + BC²

Given: In ΔPQR, ∠PQR = 90°.

To prove: PR² = PQ² + QR².

Construction:

Draw seg QS ⊥ side PR such that P–S–R.

Proof: In ΔPQR,

∠PQR = 90°   ...(Given)

Seg QS ⊥ hypotenuse PR   ...(Construction)

∴ ΔPSQ ∼ ΔQSR ∼ ΔPQR   ...(Similarity of right-angled triangles)   ...(1)

ΔPSQ ∼ ΔPQR   ...[From (1)]

∴ `"PS"/"PQ" = "PQ"/"PR"`   ...(Corresponding sides of similar triangles are in proportion)

∴ PQ² = PS × PR   ...(2)

ΔQSR ∼ ΔPQR   ...[From (1)]

∴ `"SR"/"QR" = "QR"/"PR"`   ...(Corresponding sides of similar triangles are in proportion)

∴ QR² = SR × PR   ...(3)

Adding (2) and (3), we get

PQ² + QR² = PS × PR + SR × PR

∴ PQ² + QR² = PR(PS + SR)

∴ PQ² + QR² = PR × PR   ...(P–S–R)

∴ PQ² + QR² = PR² OR PR² = PQ² + QR².

Given: In triangle PQR, ∠PQR = 90° and S is the mid-point of QR.

To prove: QR² = 4(PS² – PQ²)

in right-angled ΔPQS, by Pythagoras theorem,

PQ² + QS² = PS²

⇒ QS² = PS² – PQ²   .......(i)

Since S is the mid-point of side QR,

∴ QS = `(QR)/2`

Substituting the value of QS in equation (i),

`((QR)/2)^2 = PS^2 - PQ^2`

`(QR^2)/4 = PS^2 - PQ^2`

QR² = 4(PS² – PQ)²

Hence proved.