# Ex.6.5 Q14 Triangles Solution - NCERT Maths Class 10

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## Question

The perpendicular from $$A$$ on side $$BC$$ of a $$\Delta ABC$$ intersects $$BC$$ at $$D$$ such that $$DB$$ $$=$$ $$3CD$$(see Fig. below). Prove that $$2A{{B}^{2}}=2A{{C}^{2}}+B{{C}^{2}}$$ .

Diagram

Video Solution
Triangles
Ex 6.5 | Question 14

## Text Solution

Reasoning:

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Steps:

In $$\Delta ABC$$

$$AD\bot BC$$ and $$\,BD=3CD$$

\begin{align}BD + CD &= BC\\3CD + CD &= BC\\4CD &= BC\\CD &= \frac{1}{4}BC \ldots \left( 1 \right)\\\\&{\rm{and\qquad }}\\\\BD &= \frac{3}{4}BC \ldots \left( 2 \right)\end{align}

In $$\Delta ADC$$,

\begin{align}A{C^2} &= AD{}^2 + C{D^2}\\ & \quad \left[ {\angle ADC = {{90}^0}} \right]\\A{D^2} & = A{C^2} - C{D^2} \ldots .\left( 3 \right)\end{align}

In $$\Delta ADB$$,

\begin{align}A{B^2} &= AD{}^2 + B{D^2} \\ & \quad \left[ {\angle ADB = {{90}^0}} \right]\\A{B^2} &= A{C^2} - C{D^2} + B{D^2} \\ & \quad\left[ {{\rm{from (3)}}} \right]\\A{B^2}\,\, &= \!\! A{C^2} \! +\! {\left( \! {\frac{3}{4}BC} \! \right)^2} -\! {\left(\! {\frac{1}{4}BC}\! \right)^2} \\ & \quad\left[ {{\rm{from (1) and (2)}}}\right]\\ \\ A{B^2} &= A{C^2} + \frac{{9B{C^2} - B{C^2}}}{{16}}\\A{B^2} &= A{C^2} + \frac{{8B{C^2}}}{{16}}\\A{B^2} &= A{C^2} + \frac{1}{2}B{C^2}\\2A{B^2} &= 2A{C^2} + B{C^2}\end{align}

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