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# In ∆Abc, Seg Ad ⊥ Seg Bc Db = 3cd. Prove that : 2ab2 = 2ac2 + Bc2 - Geometry

ConceptPythagoras Theorem

#### Question

In ∆ABC, seg AD ⊥ seg BC DB = 3CD. Prove that :
2AB= 2AC+ BC2

#### Solution

It is given that,
DB = 3 CD

∴ BC = 4 CD        ....(1)

According to Pythagoras theorem,
In ∆ABD

${AB}^2 = {AD}^2 + {DB}^2$
$\Rightarrow {AD}^2 = {AB}^2 - {DB}^2 . . . \left( 2 \right)$

In ∆ACD

${AC}^2 = {AD}^2 + {CD}^2$
$\Rightarrow {AC}^2 = \left( {AB}^2 - {DB}^2 \right) + {CD}^2 \left( \text{From} \left( 2 \right) \right)$
$\Rightarrow {AC}^2 = {AB}^2 - \left( 3CD \right)^2 + {CD}^2 \left( \text{Given} \right)$
$\Rightarrow {AC}^2 = {AB}^2 - 9 {CD}^2 + {CD}^2$
$\Rightarrow {AC}^2 = {AB}^2 - 8 {CD}^2$
$\Rightarrow {AB}^2 = {AC}^2 + 8 {CD}^2$
$\Rightarrow {AB}^2 = {AC}^2 + 8 \left( \frac{BC}{4} \right)^2 \left( \text{From} \left( 1 \right) \right)$
$\Rightarrow {AB}^2 = {AC}^2 + \frac{{BC}^2}{2}$
$\Rightarrow 2 {AB}^2 = 2 {AC}^2 + {BC}^2$

Hence, 2AB= 2AC+ BC2.

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#### APPEARS IN

Balbharati Solution for Balbharati Class 10 Mathematics 2 Geometry (2018 to Current)
Chapter 2: Pythagoras Theorem
Problem Set 2 | Q: 13 | Page no. 45

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Solution In ∆Abc, Seg Ad ⊥ Seg Bc Db = 3cd. Prove that : 2ab2 = 2ac2 + Bc2 Concept: Pythagoras Theorem.
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