Geometric Mean of Similar Triangles
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Geometric Mean of Similar Triangles : The altitude to the hypotenuse of a right triangle forms two triangles that are similar to each other and to the original triangle.
ΔABC ~ ΔBDC ~ ΔADB 

Given : A right triangle ABC, right angled at B. BD ⊥ AC.
Prove that : ΔABC ~ ΔBDC ~ ΔADB


1) ∠ABC= 90  1) Given 
2) BD ⊥ AC  2) Given 
3) ∠BDC = ∠BDA = 90^{0}  3) By definition of perpendicular 
4) ∠ABD + ∠DBC = 90  4) Since ∠ABC = 90 
5) ∠C + ∠DBC + 90 = 180  5) Angle sum property 
6) ∠C + ∠DBC = 90  6) By subtraction property 
7) ∠C = ∠ABD  7) From (4) and (7) 
8) ΔADB ~ Δ BDC  8) By AA postulate ( from 3 and 7) 
9) ∠ADB = ∠ABC  9) Each 90^{0} 
10) ∠A = ∠A  10) Reflexive (common angle) 
11) ΔADB ~ ΔABC  11) By AA postulate (from 9 and 12) 
12) ∠C = ∠C  12) Reflexive (common angle ) 
13) ΔBDC ~ ΔABC  13) By AA postulate ( from 11 and 12) 
14) ΔADB ~ Δ BDC  14) from (3) and (7) 
Altitude Rule
The altitude to the hypotenuse is the mean proportional between the segments of hypotenuse.
x/h = h/y
⇒ h ^{2} = xy
Leg Rule
Either leg of the given right triangle is the mean proportional between the hypotenuse of the given right triangle and the segment of the hypotenuse adjacent to that leg.
y/a = a/c ⇒ a ^{2} = yc and x/b = b/c ⇒ b ^{2} = xc
Problem 1 : If AD = 3 and DB = 9 find CD.
Solution :
(CD) ^{2} = AD . DB
x ^{2} = 3 x 9
x = ~+mn~ √3 √9
x = ~+mn~ 3√3
CD = 3√3 ( CD can not be negative , so reject 3√3)
Problem 2 : If AD = 3 and DB = 9 find AC.
Solution :
(AC) ^{2} = AD . AB
y ^{2} = 3 x 12 ( AB = AD + DB ; AB = 3 + 9 =12)
y ^{2} = 36
y = ~+mn~ 6
AC = 6 ( AC can not be negative , so reject 6 )
Similarity in Triangles
• Similarity in Geometry
• Properties of similar triangles
• Basic Proportionality Theorem(Thales theorem)
• Converse of Basic Proportionality Theorem
• Interior Angle Bisector Theorem
• Exterior Angle Bisector Theorem
• Proofs on Basic Proportionality
• Criteria of Similarity of Triangles
• Geometric Mean of Similar Triangles
• Areas of Two Similar Triangles
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