Question

In a triangle ABC, angle
A = 90° and D is the mid point AC. The value of BC^2 - BD^2 is equal to :

Answer 1

Given : A triangle ABC
A = 90°
AD = CD
To find :
${BC}^{2} - \: {BD}^{2}$
Procedure :

Using Pythagoras theorem in ΔABD

${AB}^{2} + {AD}^{2} = {BD}^{2} ......(1)$
Using Pythagoras theorem in ΔABC
${AB}^{2} + {2AD}^{2} = {BC}^{2}.....(2)$
Subtracting (1) from (2), we get,
${BC}^{2} - {BD}^{2} = {AB}^{2} + {2AD}^{2} - {AB}^{2} - {AD}^{2}$
${BC}^{2} - {BD}^{2} = {AD}^{2}$

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